1MRB520046-Len

1MRB520046-LenEdition October 1998

Generator andTransformer Protection

Lecture

� 1997 ABB Network Partner AG / Baden / Switzerland ABB Kraftwerke AG / Baden / Switzerland

ABB Kraftwerke AG / Mannheim / Germany

3rd Edition

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ABB Network Partner AG 1MRB520046-Len Generator and Transformer Protection___________________________________________________________________

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Contents

1. Introduction ............................................................................................. 21.1. Fault Conditions...................................................................................... 21.2. Abnormal Operating Conditions.............................................................. 21.3. Basic Requirements................................................................................ 31.4. Basic Conditions ..................................................................................... 41.5. Additional Requirements......................................................................... 41.6. Overall Protection of a Generator-Transformer Unit ............................... 4

2. Currents and Voltages During Faults...................................................... 6

3. Protection Devices for Fault Conditions.................................................. 73.1 Overcurrent Protection............................................................................ 73.2 Differential Protection .......................................................................... 113.3 Minimum-Impedance Protection ........................................................... 223.4 Interturn Fault Protection ...................................................................... 233.5 Buchholz Relay or SPR ........................................................................ 293.6 Stator Ground Fault Protection ............................................................. 303.7 Rotor Ground Fault Protection .............................................................. 413.8 Transformer Ground Fault Protection ................................................... 423.9 Breaker Failure Protection .................................................................... 44

4. Protection Devices for Abnormal Operating Conditions........................ 474.1 Overload Protection .............................................................................. 474.2 Unbalanced Load Protection ................................................................ 564.3 Voltage Protection ................................................................................ 684.4 Overexcitation Protection...................................................................... 694.5 Frequency Protection............................................................................ 704.6 Loss-of-Excitation Protection ................................................................ 714.7 Pole Slipping Protection........................................................................ 754.8 Reverse Power Protection .................................................................... 804.9 Inadvertent Energization Protection...................................................... 814.10 Voltage Unbalance Protection .............................................................. 82

5. Protection Current Transformers .......................................................... 845.1. Current Transformers with Closed Magnetic Core................................ 865.2. Linear Current Transformers................................................................. 96

6. Figures and Tables ............................................................................. 100

Generator and Transformer Protection 1MRB520046-Len ABB Network Partner AG___________________________________________________________________

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1. INTRODUCTION

The task of the generator and transformer protection is maintaining operationduring:- faults- abnormal service conditions.

1.1. Fault Conditions

On electrical machines the following kinds of faults can occur and causedamage:- Short-circuits- Interturn faults- Ground faults

Most of the fault currents are of a higher level, usually higher than two-timesthe rated value (refer to the Tables); the damage is therefore great and con-tinuing operation may not be possible. Because the damage rises with time, itis necessary to interrupt operation as soon as possible by a trip of:- circuit breaker(s)- excitation circuit breaker- turbine shut down.

During stator ground faults the circulating fault currents are small; they arelimited to a level of 20 A by a grounding resistor. An arc between the faultedslot bar and the laminated stator iron causes damage to the generator andlong and expensive repair work. Therefore operation must also be interruptedby a trip. An exception are faults on ungrounded and impedance-groundedsystems having a low level of circulating currents. This is the case for:- rotor ground faults- M.V. system ground faults.With respect to restricted damage, the protection device often emits only analarm and operation is not interrupted usually.

During a short-circuit between generator and transformer (refer to Fig.1-1 or1-2) the faulted point is injected by a current of the H.V. power system and bya current of the generator.The first component can be interrupted by an ACcircuit breaker e.g. 100 ms after the fault has occured. The secondcomponent cannost be cleared by an AC circuit breaker. This component isproportional to the field current. Its interruption by a DC circuit breaker is notpossible due to a very high inductance of the field winding. It is only possibleto suppress the field current. Its suppression can be accelerated with the helpof a discharging resistor connected in parallel to the field winding.

1.2. Abnormal Operating Conditions

Under abnormal operating conditions the normal electrical, mechanical andthermal stress of protected machines is increased by


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- overcurrent- unbalanced load- overtemperature- overvoltage- over- and underexcitation- over- and underfrequency- asynchronous running- generator motoring.The protected machines are still sound, but their life can be influenced.The abnormal operating conditions could cause a fault later, if they are notdetected by a protection system in time.

The abnormal operating conditions can be caused by:- switching, failures and faults on the connected power system (they influence current, voltage, frequency)- failures on protected machines (e.g. on the cooling system)- failures on the control system (current, temperature, voltage, excitation, speed, inadvertent non- synchronized connection).

In cases of a chance to correct the abnormal conditions to normal ones, theprotection system emits an alarm (e.g. during unbalanced load, overtempera-ture etc.).

1.3. Basic Requirements

The basic requirements on the protection system are:a) Reliable operation during - internal faults - abnormal operating conditionsb) No operation during - external faults - normal operating conditions.

The high protection availability is enabled by:- the simple arrangement of important inner circuits (rule: the failure rate rises with the number of elements)- simple trip circuits- additional devices used for the same protection purpose during dangerous faults- doubling the DC voltage supply.The groundless operation of the protection is restricted by a suitable designof the internal circuits of the protection system.

In order to reinforce the protection during dangerous faults, additional devicesmay be used to increase the protection reliability. E.g. an overcurrent deviceused in addition to the differential device. At the same time the application oftwo devices for the same purpose must not considerably increase the dangerof maloperation. Therefore two devices are used usually using different


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measuring methods and having a different sensitivity. They are called mainand back-up protection devices.

1.4. Basic Conditions

Basic conditions for the correct operation of the protection system are:- suitable settings- a suitable location and specification of voltage and current transformers.

The suitable settings depend on the sufficient knowledge of minimum faultcurrents and voltages. For the specification of VT's and CT's the maximumfault currents and voltages must be applied.

The fault current level is- the highest during interturn faults- high during short circuits- very low during ground faults on generators.For calculated values of short circuit currents refer to Tables 2-III...VI.

1.5. Additional Requirements

Additional requirements on the protection system are:- selectivity- reliable operation during worst conditions.

The selectivity enables the detection of individual failures and faults. The cor-responding zone is cleared and the other system stays in operation.

The protection system must also be able to operate under worst conditions(e.g. at a reduced level of the DC voltage, failure on AVR, etc.).

1.6. Overall Protection of a Generator-Transformer Unit

With respect to safety and selectivity a number of different protection devicesare applied. Refer to the single-line diagrams in Fig. 1-1 and 1-2. During thedesign phase the following points should be considered:- Suitable split-up of the protection zones- Overlapping of zones as short as possible- Overcurrent device of the generator located at its neutral point- Unbalanced load device and reverse power device connected to CT'S and VT's with small voltage, current and phase angle errors- Voltage transformers located in the zone of the differential device- Protection devices divided into two independent groups if possible- For protection devices as those for overcurrent/undervoltage, frequency, overexcitation, underexcitation, minimum impedance, pole slipping, reverse power, only a line-to-line voltage as a supplying voltage should be used, because only this voltage is independent of ground faults- No overall differential protection should be used for units with a generator


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C.B.- Although the breaker failure protection can be provided for each circuit breaker, it is applied mainly for units with a generator C.B.

The scheme using two 52 G and 52 T circuit breakers enables the supply ofthe station transformer at an open H.V. or generator C.B. and is oftenpreferred for gas turbine generators. For steam turbine and water-wheelgenerators the 52 G circuit breaker is usually omitted.

For the REG216/316 and RET316 systems the frequency range of themeasured voltages and currents is considerably restricted by filtering for mostprotection devices. Therefore the DC component and the higher harmonicsneed not be considered during selection of the setting values.

For a suitable application of individual protection devices at different ratingsof generators and transformers refer to Tables 1-II and 1-III.


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2. CURRENTS AND VOLTAGES DURING FAULTS

The current and voltage values required for the dimensioning of the currentand voltage transformers, for various protection components as well as forthe determination of protection setting values can be calculated according tothe following figures and tables.The characteristic of short-circuit currents of transformers and generators arequite different (refer to Fig. 2-1 and 2-2).

The field-discharge circuit breaker is a modified DC circuit breaker with amain and a discharging contact (Fig. 2-3). In order to extinguish a DC currentthe arc voltage Varc of a C.B. must be higher than the voltage of theexcitation system V:

Varc > V (Fig. 2-4)

With the help of the discharging resistor the time constant T'd of the fieldcurrent can be reduced by approx. one half and the suppression of the fieldcurrent can thus be accelerated.

For symmetrical current components during asymmetrical short circuits referto Fig. 2-5, 2-6 and 2-7.

For current distribution during a line-to-line fault of transformers refer toFig. 2-8.

For current distribution during a line-to-ground fault of two transformersoperating in parallel (one unit solidly grounded and the other ungrounded)refer to Fig. 2-9.

For current distribution during asymmetrical short circuits of a generator-transformer unit refer to Fig. 2-10.

For formulas required for the calculation of currents and voltages during sym-metrical short-circuits on generators, transformers and generator-transformerunits refer to Tables 2-I to 2-VI.


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3. PROTECTION DEVICES FOR FAULT CONDITIONS

3.1. Overcurrent Protection

The overcurrent protection is used:- to supervise service overcurrents- to detect faults as a back-up protection.

The overcurrent protection for generators and transformers is different due todifferent requirements given by the different characteristics of the short-circuitcurrent (Fig. 2-1 and 2-2).

3.1.1. Overcurrent Protection for Generator-Transformer Units

3.1.1.1. Definite Time Overcurrent Protection

The task of this protection device connected to CT's located at the gener-ator neutral point is to protect:a) the zone between the generator neutral point and the H.V. circuit

breaker against service overcurrentsb) the zone from the generator neutral point up to the H.V. power system

as a back-up protection during faults (If a M.V. bus is applied, it is also involved in the protected zone).

The current curve of a suddenly shorted generator comprises dampedsubtransient and transient components (Fig. 2-2). After a certain time ofseveral seconds the overcurrent measuring system drops out, becausethe instantaneous current values do not exceed the current setting levelanymore.If the A.V.R. is not active the continuous steady-state short-circuit currentis smaller than the rated stator current, due to the high value of the syn-chronous reactance. At xd = 2 the continuous current i = 0.5. It is evenzero when the excitation system is supplied from the generator terminals.

The definite time overcurrent characteristic with a desired delay of severalseconds would prohibit the trip at such a curve of the fault current.

For satisfactory tripping a combination of an overcurrent and an undervol-tage measuring system (ANSI device Number 51 V, Fig. 3.1-2) is to beused.

The current setting must be chosen between the minimum through-faultcurrent (usually during a line-to-line fault on the H.V. side) and themaximum short-time overcurrent (Fig. 3.1-1).

This maximum short-time overcurrent is to be determined in accordancewith the:- maximum continuous operating current


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- switching conditions in the H.V. power system- influence of the ceiling voltage- accuracy of current detection- reset ratio.

An example of calculated short-circuit currents supplied by a generator-transformer unit in Table 2-VI shows that the highest short-circuit currentcirculates during a line-to-ground fault. During H.V. faults the currentcirculating on the generator side has the following consequences:- During a line-to-ground fault the through-fault current is 2.89 p.u.- During a line-to-line fault the through-fault current of one phase is 4.0 p.u. and 2.0 p.u. of the other phases.

The minimum through-fault current is therefore equal to 2.0 p.u. and cir-culates through the generator CT's during an H.V. line-to-line short-circuit.

A suitable current setting is usually i = 1.5. The undervoltage setting valuemust be higher than the maximum line-to-line voltage measured on thegenerator terminals during all possible kinds of faults on the M.V. and H.V.side. The usual voltage value is v = 0.7.

The value of the applied delay depends on the grading with overcurrentdevices of the power system and of the auxiliaries. The setting value ofthe delay is usually in the range between 2 and 5 s. At its selection thepermitted maximum through-fault current duration of 2 s for many trans-formers is to be considered.

3.1.1.2. Instantaneous Overcurrent Protection

Sometimes an instantaneous stage is asked for. This stage should enablefast clearing of short-circuit currents during faults between the generatorC.B. (if applied) and the generator transformer.

For selective protection the current value setting is to be chosen in therange between:- minimum short circuit currentsand- maximum through-fault currents during faults on the H.V. side.

The minimum short-circuit fault current would be a line-to-line fault. Themaximum current during an H.V. fault circulates during a three-phaseshort-circuit.

If, for example, we use the following reactance values

xd'' = 0.15 x2 = 0.15 xT = 0.1

we get


9

i 0.9 3x x

0.9 30.15 0.15

5.2d

"2

�

�

�

�

�

for the minimum fault current (at a voltage of 90%)and

i= 1.1x x

1.10.15 0.10

1.10.25

4.4d

"t�

�

�

� �

for the maximum through-fault current during an H.V. three-phase fault (ata voltage of 110%)

For protection devices with a significant restriction of the influence of theDC component, a possible setting would be i = 4.8 or 5. Without that res-triction, an H.V. fault could cause a trip of the generator C.B. In this case itis better to omit the instantaneous stage.

For large machines the difference between the minimum fault current andthe external maximum fault current is smaller due to a higher value of xd''.It is then even more difficult to find a suitable setting level.

If the generator-transformer unit has no generator circuit breaker, the in-stantaneous overcurrent device connected to CT's located at thegenerator neutral point would not make any sense.

3.1.2. Overcurrent Protection for Generator Transformer

The application of a generator circuit breaker enables the supply of theauxiliaries from the H.V. system by the generator transformer at an opengenerator C.B. In this case the generator transformer should be protectedby its own overcurrent device aside from the other protection devices. Thisis a usual arrangement for gas turbine generators.

The preferred type of the transformer overcurrent protection has a definitetime overcurrent device and often also an instantaneous overcurrentdevice. This overcurrent protection is afforded by H.V. currenttransformers. The short-circuit current curve of transformers has nodamping of the AC component (refer to Fig. 2-1). Therefore no additionalvoltage monitoring is desired and only an overcurrent measuring device isused.

3.1.2.1. Definite Time Overcurrent Protection

The task of the definite time device is a selective detection of short circuitson the transformer and on its M.V. side supplied by the H.V. powersystem. This device must be graded with the overcurrent devices of theM.V. side of auxiliary transformers. Naturally this device must not operateduring continuous operation current as well as during short-time


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overcurrents. This device must operate at all short-circuits on the M.V.side. The minimum fault current circulates during a line-to-line fault; itsvalue is 8.66 IN at the transformer reactance of 10% (refer to Table 2-V).If we assume a maximum short-time overcurrent level of 1.3 IN then thepossible setting range is between 1.3 IN and 8.66 IN.

3.1.2.2. Instantaneous Overcurrent Protection

The task of the instantaneous overcurrent device is the protection againsthigh current faults on the transformer H.V. side supplied by the H.V.power system.

The instantaneous device must operate during H.V. faults between theH.V. current transformers and the transformer H.V. winding. It must notoperate during M.V. faults and transformer inrush currents. Theapplication of an instantaneous device is especially important at longdistances between the H.V. current transformers and the transformerH.V. winding, threatened by a very high fault current level. E.g. if the H.V.current transformers are located close to the H.V. circuit breaker at thebegin of a line supplying this transformer.The current setting must beselected between- minimum fault current of the H.V. sideand- maximum through-fault current.Designating the H.V. power system reactance xS and the transformerreactance xT, the minimum H.V. fault current is

i 32xF

S�

during a line-to-line fault.The maximum through-fault current is

i 1xT

�

during a symmetrical three-phase fault.

Let us assume: xS = 0.02; xT = 0.1 and we get

ix

iF � � � �

32 0 02

43 10 1

10. .

A sensible current setting would then be e.g. i = 15 (Fig. 3.1-3).

In the case that an inrush current is higher than the through-fault current,the bottom limit of the instantaneous stage setting range is to be specifiedby the inrush current level instead of the through-fault current level. Very


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high inrush currents occur at transformers supplied by a power systemwith a very high level of short-circuit currents.

For the REG216/316 and RET systems the instantaneous stage can beblocked if the circulating overcurrent has a 2nd harmonic componentwhich is higher than 10 % of the current setting level.

3.1.3. Overcurrent Protection of Auxiliary Transformer

Overcurrent protection of the input side:The protection device comprises a definite time stage and if required alsoan instantaneous stage like the generator transformer.

Overcurrent protection of the output side:This protection device has only a definite time stage; it is to be gradedwith other overcurrent devices of the connected and supplied auxiliarysystem.

For application of overcurrent devices refer to Fig. 3.1-6. For an exampleof grading with possible current and delay setting values refer to Fig.3.1.7.

3.2. Differential Protection

3.2.1. Introduction

The differential protection is the main and most important protectionagainst short-circuits. Transformer interturn faults and ground faults ofsolidly grounded windings are additionally detected. As a back-upprotection a minimum impedance overcurrent device and a Buchholz or asudden pressure relay (SPR) are used.

The differential protection is used for the detection of internal faults withina zone defined by the location of the supplying current transformers. Itoperates with a comparison of the currents before and behind theprotected machine (Fig.3.2-1). The differential protection must besensitive and fast (Fig. 3.1-2). Otherwise the damage caused by high faultcurrents, rising with time, would be too high. The high sensitivity isdemanded in order to detect most of the faults; this is important especiallyduring transformer ground faults. It is usually accepted that the differentialprotection detects failures of the secondary CT circuits (interturn faults,short-circuits as well as interruptions) and trips. During external faults thedifferential protection must not trip (Fig. 3.2-1).


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3.2.2 Differential Protection for Transformers

3.2.2.1. Application

The differential protection for transformers must be designed to meetthree conditions:- sensitive detection and fast clearing of internal faults- sufficient stability during external faults- sufficient stability at transformer inrush currents (Fig. 3.1-4).

To satisfy these requirements, the differential device for transformerscomprises:- a sensitive instantaneous overcurrent system- a sensitivity restraint by the bias circuit- a sensitivity restraint acc. to the 2nd harmonic current.

The sensitivity of the differential overcurrent detection during internalfaults is set by a basic setting "g" for ABB relays. The stability of thedifferential protection during external faults depends on the factor "v"defining the slope of the trip characteristic.

3.2.2.2. Basic Setting "g"

The sensitivity of differential devices during internal faults is given by thebasic setting value "g". This value should be selected as low as possiblein order to enable the detection of most of the faults but withoutjeopardizing any device operation at- Generators due to CT current errors- Transformers due to

- CT current errors- transformer no-load current (especially at short time overvoltages)- tap changer operating range.

For an average magnetizing curve of transformers refer to Fig. 3.2-3.

Using the protection class 5P20 for CT's the differential current causedby the current error of CT's is usually considered to be 10 %.

The transformer no-load current depends on its design. Nevertheless acc.to Tabel 3.2-I a value of 10 % can be used if no actual value is known.

The differential current caused by the tap changer depends on its range,which can be e.g. +/- 5 % or +/- 10 %.

If we consider the sum of these 3 components to be 30 % as the worstcase, then the minimum setting value "g" should be 0.3 for transformers(Tabel 3.2-I).


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3.2.2.3. Factor "v"

The stability of the differential protection depends on the slope of the ope-rating characteristic given by the factor "v" (sometimes called "pic-upratio"). The value "v" is given by the ratio of the pick-up differential current� I to the through current IH (Fig. 3.2-4).

The value "v" should be selected such that:- during normal operation also small fault currents can be detected- danger of a false tripping during external faults due to CT saturation is restricted.

The ABB differential protection uses an operating characteristic, which isa combination of a constant curve and of two slope curves. Theprotection stability is influenced during:- through-currents of the medium level by the gentle slope- high through-currents by the steep slope.The gentle slope can be set by the ratio "v". The steep slope is fixed.The danger of false tripping during external faults is considerablyrestricted by the steep slope.

For CT's with a different behaviour at off-set overcurrents, higher values of"v" are to be selected. The normal setting value is v = 0.5.

At the REG and RET protection systems the steep slope is activated onlyduring CT secondary currents I > b IN, where "b" is a certain throughcurrent in p.u. By setting the value "b" the point of intersection of bothslopes is selected. The steepness of the steep slope is infinite (Fig. 3.2-4).The usual setting value is b = 1.5. In the case of danger of a falsetripping at low through currents the value b = 1.25 can be applied.

3.2.2.4. Matching and Filtering

Very often rated currents of the protected transformer and of the CT's aredifferent.

The difference between the ratio of the transformer rated currents and theratio of CT's rated currents must be matched. Otherwise this differencewould produce a spurious differential current. This matching is possible forthe REG216/316 and RET316 systems with the help of "referencevalues".

It is an advantage if the setting value "g" (p.u.) can be based on the ratedcurrents of the transformer and not of the CT's. The phase shift betweenthe currents of the primary and secondary transformer side is given by theconnection group and must also be matched.


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An external line-to-ground fault of a Dy-connected transformer causes acirculating current with positive, negative- and zero-sequence componenton the faulted "wye" side. The line current supplying the "delta" side hasonly a positive and negative-sequence component. The zero-sequencecurrent circulates only in the delta-connected winding. The differentialdevice measuring the line currents on the primary and secondarytransformer side, would detect an unbalance, if the zero-sequencecomponent of the"wye" side is not prohibited to circulate to the differentialdevice. Therefore the following remedies are applied for the transformerdifferential protection:- Current matching- Connection group matching- Filtering of the zero-sequence component.These remedies can be performed by a suitable current ratio and acorrect connection group of interposing CT's or by modern protectionsystems alone.

3.2.2.5. Increased Basic Setting g-High

The REG and RET systems offer the possibility to reduce the sensitivityby increasing the value "g" ,if it is required, with respect to- increased transformer no-load currents (due to a short time overvoltage)- differential currents caused by a tap changer- various other purposes.

A suitable setting must be selected according to the purpose.

3.2.2.6. Differential Current I-Inst

The operating time of the transformer differential device is independent forthe REG216/316 and RET316 systems on the energizing detector duringdifferential currents exceeding the setting value I-Inst. Such a shorteroperating time is important during internal faults with a very high currentlevel.

3.2.2.7. Transformer Inrush Current Detection

The inrush current detection operates with the detection of the 2ndharmonic. During normal operation this detection is not activated. It can beactivated by the:- energizing detector- external signal "InrushInp".

The energizing detector operates when the setting value of the "inrushratio" is exceeded, i.e. when the content of the 2nd harmonic exceeds thesetting value.


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The inrush current detection is only active during the "InrushTime", whichmust be set.

3.2.3 Differential Protection for Generators


The differential protection for generators must be designed to meet twoconditions:- Sensitive detection and fast clearing of internal faults- Sufficient stability during external faults.

The inrush current of newly energized transformers can also circulate inthe generator stator winding; but for generators this is a through-currentand no differential current such as for transformers. Therefore thesensitivity restraint for inrush currents is omitted at some differentialprotection devices for generators, such as e.g. at the REG216 and 316systems. Also matching and measuring of the 2nd harmonic are omittedfor these systems.

3.2.3.2. Settings

The highest applicable sensitivity is limited by different current errors ofboth CT's; therefore the minimum setting value "g" should be 0.1(p.u.).

The factor "v" is usually set to the value 0.25 with respect to not especiallyhigh through-currents during external faults. If necessary the higher valueof 0.5 can be applied.

The factor "b" cannot be set at the differential device of the REG216/316systems for generators. Its value 1.5 is fixed.

3.2.4 Suitable CT's for Differential Protection


The current transformers applied for the differential protection mustenable correct operation of the protection during internal as well as atexternal faults.

For generator-transformer units the differential protection must be stableduring H.V. faults (Fig. 3.2-6a). The fault current level is limited byreactances of the generator and of the transformer. Similarly the timeconstant of the through fault current is given by the time constant of thegenerator and of the transformer.

The units for gas turbines have a generator circuit breaker which en-


16

ables the supply of the auxiliaries also at a disconnected generator. In thiscase the transformer differential protection must be stable during short-circuits in the generator zone supplied from the H.V. side and enable thesupply of auxiliaries without any service interruption (Fig. 3.2-6b).Naturally the generator differential protection must also be stable duringfaults on the auxiliary feeders (Fig. 3.2-6c).

The CT's designed for conditions of external faults mostly enable correctprotection operation during internal faults too. Therefore the CT's areusually specified with respect to the protection stability during externalfaults. This stability depends on the behaviour of the protection device andof the CT's. The differential protection device has a suitable operatingcurve as a means for device stability. Therefore it is not necessary todemand that CT's do not saturate. Such a requirement would cause bigCT's with a high magnetic core cross section.

Nevertheless the applied CT's must be able to transfer such a current tothe CT secondary side, which enables correct operation of the differentialprotection device.

The ability of CT's to transfer overcurrents to the secondary side withoutany saturation is specified by the overcurrent factor "n". Most of theapplied CT's are not specified for the full value, but for a reduced factor"n" with respect to the stabilizing form of the operating curve of thedifferential device.

The current transformers designed for protection purposes are specifiedaccording to the class and the rated output. CT's specified according toIEC standards are usually of the class 5P20 . The maximum current errorallowed is then 5 % at a current 20 times the rated value. The currenterror is the difference between the primary and secondary current, if bothcurrents are based on the same number of turns for the primary andsecondary side (Fig. 3.2-5).

The factor 20 called "overcurrent factor" is used to specify the highestcurrent, during which the current error is still acceptable for the operationof protection devices.

3.2.4.2. Calculation of the Overcurrent Factor

The overcurrent factor "n" desired for a faulted circuit and non-saturatingCT's is defined by a continuous AC overcurrent without any DCcomponent.

This factor "n" may also be used for offset overcurrents having an ACcurrent component "i" and a DC component. The influence of the DCcomponent is specified by a factor "k" :


17

n = k i

The factor "k" is a function of:- frequency- time constant of the faulted circuit �- time constant of the CT secondary circuit �2- time t.

The factor "k" is determined by the exponential functions given by theshort-circuit and by the CT secondary circuit. It depends on the instan-taneous voltage at the instant of the short-circuit.

Short-circuit at voltage v = vmax :In this case the short-circuit current has only an AC component.

iI 2

1

1N� sin �t

The CT secondary current curve thus comprises an induced ACcomponent and an exponential function of the CT secondary circuit. Thefactor "k" is given by the CT magnetizing current and is

k cost

� �

�

e t�

�2

The maximum value of "k" appears at � �t � :

k = 2

Short-circuit at v = 0:The short-circuit current has a DC and an AC component.

iI 2

cos1

1N

t

� �

�

e t�

�2

Assuming a CT linear magnetizing curve, the factor "k" is then(Fig. 3.2-7):

k e e 1��

��

��

�

��

��

��

�

� �

��

2

2

2

t t

The first formula for the factor "k" can be used:- at fault currents without any DC component- if the influence of a CT saturation, due to the DC component, can be accepted.


18

The value k = 2 can be used e.g. for ground faults on resistancegrounded systems, where the fault current has almost no DC component.

The second formula is to be used if the CT must not saturate during anoff-set fault current. The function of "k" given by two exponentialfunctions has a maximum value at a certain time (Fig. 3.2-7):

tm n��

�

��

�

��

� �

� �

�

�

2

2 2

�

k 1m ��

��

�

��

��

�

�

�

� �

2

2

The specification of a CT according to "k" makes sure that the CT neversaturates under given short-circuit conditions. The calculation for the faultinstant at v = 0 is the worst case. If it is accepted that a mechanicalreason for a fault is not probable and that an electric break-down must beinitiated by a certain voltage, then the result of the calculation of "k" acc.to the second formula can be reduced. E.g. a reduction of the DCcomponent and also of the factors "k" by 0.866 corresponds to aninstantaneous voltage of 50 % of the peak value.

Current transformers with a magnetic ring core have a high time constant�2. If the " �2 " is much higher than " � "

� �2 ��

an approximate formula for "k" can be used:

kt

� ��

��

�

��

�

��1 1e

3.2.4.3. Current Transformers Specified acc.to IEC Standard

Protection current transformers of the class 5P20 have a magnetic coredesigned for an inner voltage E2 given by 20 times the value of the ratedoutput voltage V2N and by 20 times the value of the voltage drop R2 I2Nacross the resistor R2 (Fig. 3.2-5). The magnetic core of CT's forgenerators and many transformers is a ring core, at which the leakageinductance of the secondary winding is very small and may be neglectedat the calculation:

E2N = V2N + R2 I2N


19

E2 = 20 ( V2N + R2 I2N )

The internal voltage E2 is the maximum voltage produced during currenterrors < 5 %. The same CT may be used when a higher overcurrentfactor "n" is required and if the actual burden is smaller:

E2 = n ( V2 + R2 I2N )

If we compare the last equations we get:

n V R IV R I

N N

N�

�

�20 2 2 2

2 2 2

or using power instead of voltage:

rated output SN = V2N I2Nsecondary copper losses � PCu2 = R2 ( I2N )2actual output S = V2 I2N

n 20 S PS P

N Cu2

Cu2

��

�

�

�

This equation shows that an overcurrent factor "n" higher than 20 can bereached if the actual burden is smaller than the rated burden at SN. Theabove formula is to be used if a certain CT is to be applied for a requiredhigher factor "n" . In opposition if the desired overcurrent factor "n" wascalculated, the actual burden is known and the CT copper losses can beestimated, the CT is to be specified for class 5P20 by the calculatedrated output

� �Sn

20S P PN Cu2 Cu2� � ��

such as e.g. class 5P20 and rated output SN = 90 VA.

Example:

Required value n = 40Actual output S = 5 VASecondary copper losses �PCu2 = 35 W = 35 VA

Using the equation

� � � �Sn

20S P P

4020

5 35 35N Cu2 Cu2� � � � � ��

S VAN � 45


20

3.2.4.4. Current Transformers Specified acc. to ANSI / IEEE Standard

Protection CT's according to ANSI or IEEE and having a magnetic ringcore (a low inductance type) are specified by the class C and by theoutput voltage "V2" at 20 I2N .

E.g. the class C 400 means a ring CT with an output voltage of 400 V at20 I2N and of 20 V at I2N . If the rated output current is 5 A, the ratedoutput would beSN = 20 x 5 = 100 VA.

The CT's specified acc. to ANSI or IEEE standard are designed for aninner voltage E2 given by the output voltage and by the voltage dropacross the resistor R2. The allowed maximum current error is 10 % at 20I2N .These CT's may also be applied for a higher overcurrent factor if theactual burden is smaller than the rated burden.

3.2.4.5. Requirements on CT's Applied for the REG and RET Systems

There is no simple method of calculating the performance of thedifferential device together with its CT's under fault conditions. Apossibility is an investigation of that performance by measurement atsimulated fault conditions.

The performance of the differential functions of the RET316 systemtogether with supplying CT's was investigated in the range between 3 INand 20 IN with and without a DC component given by the time constantbetween 40 and 300 ms. The results show the following:

a) Considering the stability during external faults under above conditions:The CT's may saturate and the stability is unaffected at the followingminimum overcurrent factor "n" values:- between 5 and 15 for transformers with 2 windings- between 9 and 33 for transformers with 3 windings.

b) The operating time during internal faults under above conditions isunaffected by fault currents if the inrush detector is not activated.

The above results for transformers with 2 windings can also be used forgenerators. Thanks to the similar design all above results can also beused for the REG216 and 316 systems.

Conclusion:- CT's of the class 5P20 may be used for the differential protection devices for generators and 2-winding transformers- CT's of the class 5P20 may also be used for the differential protection

of 3-winding transformers if the rated output SN is calculated with respect to the required factor "n" and to the actual required output S;


21

i.e. if the rated output SN is higher enough than the required output S in order to enable the required value "n".

3.2.4.6. Trouble shooting

Difficulties during the operation of the differential protection can be causedeither by the differential protection or by the current transformers.

A very good possibility of checking the function of the differential deviceunder normal operating conditions is the measurement of the differentialcurrent.An unusually high differential current can be caused by:

- opposite polarity of CT's- wrong current matching- wrong vector group- harmonics.

False tripping of an energized transformer at no-load can be caused by aninsufficient inrush current detection. Check the setting "InrushRatio" and ifnecessary the content of the 2nd harmonic in the curve of the measuredinrush current.

False operation during external faults can be caused by CT saturation aswell as by an unsuitable setting especially of the value "v". The differentialcurrent produced by different CT secondary currents can be caused by:

- different current ratioes- unadapted burden- CT no-load curve (magnetizing curve).

Under the assumption of a correct CT turn ratio, the accuracy of the CTcan be checked with the help of several points of the CT magnetizingcurve. Supplying the secondary side at a primary side with I1 = 0 themeasured current is a no-load current, producing the current error of theCT. The supplying voltage is to be adjusted to the internal voltage E2 .This value is given by the sum of the terminal voltage at 20 I2N and ofthe voltage drop across the resistor R2 at 20 I2N at the IEC class 5P20as well as at the ANSI or IEEE class "C". The measured current valuemust not exceed 5 % of 20 I2N , i.e. I2N , according to IEC or 10 % of 20I2N , i.e. 2 I2N , according to ANSI / IEEE class "C" (Fig. 3.2-8). At thevoltage E2 given by the rated current I2N the measured no-load currentmust not exceed 1 % of I2N .

3.2.4.7. Checking the Differential Devices

Aside from the internal simulation of a differential current at the REG216,316 and RET316 systems, an injected current or a generator current canbe used for the test. With the help of an injected current at machinestandstill the following can be measured:


22

- the value "g"- the operating timeand with the help of the running generator, supplying a symmetrical shortcircuit arranged by a link, the following can be checked:- protection operation during an internal fault- protection stability during an external fault.

If it is not possible to arrange an artificial internal fault by a link or bygrounding isolators, an artificial differential current can be producedshorting an input current circuit.

In any case the artificial short-circuit must be symmetrical; otherwise thegenerator rotor could be damaged.

3.3. Minimum Impedance Protection

This protection is a back-up protection of the generator-transformer unitagainst short circuits. For the protected zone refer to Fig. 3.3-1. Theadvantage of this protection function is a shorter tripping time than thetime required by overcurrent functions.

The minimum impedance device is connected to voltage transformerslocated at the generator terminals (Fig. 3.3-1). The device is usually set toprotect the connection to the generator transformer and its M.V. winding.The H.V. winding cannot be well protected by this protective device due tothe danger of maloperation for faults on the H.V. side.

A short delay is to be used to avoid any maloperation. This is especiallythe case for units with a generator C.B., due to:- grading with the main protection (differential protection)- transformer inrush currents.

A maloperation would also be caused by the loss of a VT voltage; in thiscase the minimum impedance function must be blocked.

The impedance setting is selected according to 70% of the transformerimpedance voltage. Assuming the transformer impedance voltage to be10% and the same rating of the generator as well as of the transformer weget for- impedance setting 0.07 p.u.- usual delay values at

- units without generator C.B. 0.2 s- units with generator C.B. switching the transformer (inrush currents) 0.5 s


23

3.4. Interturn Fault Protection

3.4.1 Interturn Fault Protection of Generators

The interturn faults of generator and motor stators cannot be detected bythe differential protection, because the measured currents at both sides ofthe winding are the same. Therefore other detection methods must beused.

The ABB generator protection scheme uses detection methods monitoring: - Voltage (Fig. 3.4-1 - Current (Fig. 3.4-2,3) - Differential current (Fig. 3.4-4).

All above methods use the decreased magnetic flux of the faulted phasefor the detection. A large current circulating in the shorted turn or turnsproduces a magnetic flux of the opposite polarity in contrast to the polarityof the original flux. The total magnetic flux is then much smaller and thephase voltage too.

At a winding having two circuits connected in parallel per phase an interturnfault causes a circulating current which can be monitored.

Features of the voltage method - simple, sensitive - 3 VT's with both primary terminals insulated for the high voltage - applicable sensitivity limited by - asymmetry of the 1st harmonic of phase voltages - zero-sequence voltage caused by the 3rd, 9th and other harmonics.

Features of the current method: - useful only for windings having two circuits connected in parallel - applicable sensitivity limited by unsymmetry of voltages of windingcircuits connected in parallel, - 1 or 2 CT's.

Features of the differential method: - applicablel only for windings having two circuits connected in parallel - applicable sensitivity limited by unsymmetry of voltages of winding circuits connected in parallel, - 6 CT's - monitoring of the circulating current individual for each phase.

The differential method is the sole one enabling the detection of the faultedphase.


24

This protection should be fast to restrict the damage caused by high faultcurrents. Nevertheless sometimes a short delay between 0.1 and 0.5 s maybe necessary to avoid any maloperation. This especially holds true for themethod using the voltage measurement.

3.4.2 Interturn Fault Protection of Transformers

The interturn fault of a transformer may under certain circumstances be de-tected by the differential protection using the comparison of primary andsecondary currents, contrary to generators and motors at which thecurrents are compared at the begin and the end of the winding and do notenable any detection of interturn faults. Some of the interturn faults canalso be detected by an overcurrent protection scheme.

The transformer currents caused by an interturn fault depend on thenumber of shorted turns, on the transformer leakage reactance and on thecircuit resistance. The leakage is variable, depending on the number ofshorted turns.

Interturn faults across a smaller number of winding turns may cause smalland not well measurable circulating currents. Such interturn faults shouldbe detected by a gas protection device like the Buchholz relay or the SPRdevice. Nevertheless, in the range of interturn faults with more shortedturns the currents are high enough to be detected by a differential or anovercurrent protection.

3.4.3 Voltages and Currents at Interturn Faults

3.4.3.1 Interturn Faults on Generators

Any interturn fault on a phase of the generator stator winding causesa reduced magnetic field of that phase. The consequence is a reducedphase voltage. A supervision of the change of this voltage enables thedetection of interturn faults (Fig.3.4-1). During interturn faults on windingswith branches connected in parallel the current distribution in thosebranches is changed and the current circulating between those branchescan be used for the fault registration (Fig. 3.4-2,3,4). The sensitivity of theinterturn fault detection depends on the ratio of sound and faulted turns ofthe failed branch and naturally on the protection setting.

The level of the voltage displacement depends on the - number of shorted turns - kind of service of the generator.


25

In an extreme case the total length of a stator phase winding is shorted; i.e.all turns of a phase winding are shorted and the voltage on this faultedphase fully disappears. The two different service kinds are: - generator disconnected from the power system (running at no-load) - generator directly connected to a large power system

At no-load the potential of the terminal of the faulted phase moves to theneutral point. The line-to-line voltage between the sound phases is fixed,non influenced. The line-to-line voltages between the faulted phase and 2sound phases drop to the phase voltage value. Using the measurement ofthe neutral point voltage displacement by the broken delta side of threeVT's connected in parallel with the generator phase windings the voltagemeasured during an interturn fault along the total phase winding length is1/3 of the maximum value e.g. of 100 V (Fig. 3.4-5a).

On a generator connected to a fixed voltage of a large power system allthree line-to-line voltages are fixed. The voltage measured across thebroken delta of VT's rises with the number of short-circuited turns up to themaximum value e.g. of 100 V at an interturn fault across the total phasewinding length (Fig. 3.4-5b).

Therefore the sensitivity of the interturn fault detection by the voltagemeasurement is smaller at the disconnected generator.

The detection of interturn faults by the supervision of the voltage accrossthe broken delta of VT's connected to the generator terminals isindependent of ground faults. On the other hand a supervision of thegenerator neutral point displacement, used for the detection of statorground faults, is not only able to detect ground faults but also interturnfaults.

In the case of a stator winding with 2 or more branches connected inparallel for each phase, it is preferable to apply a circulating current or adifferential current for the interturn fault detection. These currents dependon the ratio of the shorted turns to the total number of all turns per phaseconnected in series and on the number of parallel branches.

Currents in a disconnected generator (Fig. 3.4-6): - On a winding without branches connected in parallel the current can circulate only in the shorted turn(s) - On a winding with branches connected in parallel

- a high current circulates in the shorted turn(s)- a smaller current circulates between the faulted and the sound branches.

Generaly during interturn faults the currents circulating on a winding with2 branches connected in parallel can be calculated according to thefollowing equations developed from the circuit in Fig. 3.4-7.


26

The current i1 circulating in the sound section of the faulted branch:

i = 1x

N N

N + N 2 N N 1 N1

3

2

2 2 3

1 12

3 32

1 32

22

�

�

�

� �

�

�

� �

The current i2 circulating in the section with the interturn fault:

i = NN

1x

1

1 + N

N + N 2 N N N

23 3

2 2 2

2

22

1 12

3 32

1 3 22

�

� �

� � �

�

� �

The phase voltage u1 reduced by the influence of the interturn fault:

� �u1

1 2 1

3 3

2 1 32

22

1 12

3 32

1 32

22

= UE =

NN 1 +

N N N + 1

N

N + N 2 N N 1

N

1�

�

�

�

� �

�

�

� �

�

�

��

�

�

��

The measured voltage change �u1:

� u = 1 - u1 1

Used symbols:E E.M.F. per phase, induced from the rotorU1 , u1 phase voltage during an interturn fault�u1 phase voltage change caused by an interturn faulti1 current of the sound section in p.u.i2 current of the section with the interturn fault in p.u.N1, N2, N3 number of turns of section 1, 2 and 3�1, �2, �3 Blondel's factors (ratio of the total flux to the excitation flux,

defined by the leakage)�1 , �2 , �3 leakage factorsx subtransient reactance.

From the diagram im Fig. 3.4-8 it is evident that the minimum number ofshorted turns, which can be registered, depends on the ratio of faulted andtotal turns per phase and naturally on the protection voltage setting.Therefore a better detection sensitivity is reached at turbogenerators whichhave a smaller number of turns per phase as e.g. N3 =10, in comparisonwith salient pole generators having e.g. N3 = 40.

In the absence of parallel branches we get the simple formulas from theabove modified equations:


27

i1 = 0

i = NN

1x2

3 3

2 2

�

�

u = N N1

2 1

2 3

�

�

The last formula shows that the reduction of the phase voltage depends onthe leakage factor �2. Without this leakage the voltage u1 would sink tozero.

Generator connected to the power system: - The currents not only circulate in shorted turn(s) and between thefaulted and sound branches of the faulted phase but also between thegenerator and the power system.

In this case the currents circulating in the faulted winding are overlappedwith currents circulating from the power system.

3.4.3.2 Interturn Faults on Transformers

The Interturn faults on transformers can sometimes be detected by thedifferential protection, contrary to generators and motors at which thecurrents are compared at the begin and at the end of the winding and donot enable any detection of interturn faults.

The transformer currents during an interturn fault depend on thetransformer leakage reactance and on the circuit resistance. The leakage isvariable, depending on the number of shorted turns.

The current curves for the following diagrams were calculated under theassumption of a certain variability of the leakage reactance.

Interturn Faults on the Transformer Secondary Winding

The circulating currents can be calculated acc. to the following formulas:

i = 13

1

23

x + x 5 4NN + r

NN

1

T2sc

2

22

2

2 2

ssc

��

��

�

��

�

�

�

�

��

�

��

�

�

�


28

i = i 3 N

N2sc 12

2sc

Used symbols:i1 primary current in p.u. of the transformer rated currenti2sc current in the interturn faulted winding section in p.u.xs reactance of the supplying power system in p.u. of the transformer

ratingxT transformer reactance (impedance voltage) in p.u.r resistance of the faulted circuit in p.u.N2 number of secondary winding turns (per phase)N2sc number of shorted turns of the secondary winding.

With respect to the protection sensitivity desired for the fault detection, thecurrents circulating during interturn faults were calculated considering thesevere conditions given by the great variability of the transformer leakagereactance being between 10 and 50% and by the circuit resistance.

Fig. 3.4-9 shows that during interturn faults on a secondary winding with aconstant leakage reactance and without any resistance, the primary as wellas the secondary currents are high enough for the detection at eachinterturn faulted winding length "x". The leakage reactance variabilitycauses the circulating currents to be smaller. For the influence of the circuitresistance refer to the (Fig. 3.4-10). It is shown there that especially therange of short faulted winding lengths is influenced by the circuitresistance. Nevertheless in the range of the longer faulted winding lengths(above 20% of the total winding length), the measured primary currents arehigh enough to be detected also by a simple overcurrent protectionscheme.

Conclusion:In the range of the faulted secondary winding lengths between 20 and100% the currents are influenced especially by the variability of thesecondary winding leakage. In the range between 0 and 20% the currentsare restricted especially by the circuit resistance.

Interturn Faults on the Transformer Primary Winding

The circulating currents can be calculated acc. to the following formulas:

� �

i =1

3 r x x

1

2+ 2 s + 3

N1 N1scN1

2 1 5 4

N1scN1

2��

��

�

��

�

�

��

�

�

��

�

�

��

�

�

��


29

i = i N

N1sc 11

1sc

Used symbols:i1 primary current in p.u. of the transformer rated currenti1sc current in the interturn faulted winding section in p.u.xs reactance of the supplying power system in p.u. of the transformer

ratingx1� primary leakage reactance of the transformer in p.u.r resistance of the faulted circuit in p.u.N1 number of primary winding turnsN1sc number of shorted turns of the primary winding

For interturn faults on the transformer primary winding the critical range forthe detection is again at the short interturn faulted winding lengths(Fig. 3.4-11,12), especially due to the circuit resistance. Nevertheless themeasured primary current value is often sufficient for the detection also inthis range.

3.5. Buchholz relay

The Buchholz relay is a gas detection relay protecting oil transformersduring- internal insulation faults such as short circuits, interturn faults, earth faults- small failures such as spark emission, discharges, iron lamination faults- tank leakages.

The Buchholz relay supervises gas emissions and is built into the oil pipeconnecting the transformer tank with the expansion chamber. The relayconsists of a small reservoir to accumulate the incoming gas, and of twoswimmers to monitor the oil level located one above the other.

The relay has two stages for an alarm and a trip. The upper swimmeroperates the very sensitive 1st stage protection and is able to detect smalloil level differences caused by small gas emissions or by a small oilleakage.

The lower swimmer operates the 2nd stage protection; it detects greater oillevel differences and oil stream surges caused by sudden insulationfaults.The second stage detecting the fast oil streams caused by insulationfaults trips in approximately the same time as the differential protection.

The Buchholz relay is a second protection device against heavy faults. Ithas a quite different detection system than the differential device. Itsadvantage is a very high sensitivity and ability to detect faults and failureswhich cannot be detected by the differential device. The zone protected


30

by the Buchholz relay is limited to the transformer tank, whilst the zone ofthe differential protection is given by the location of current transformersinstalled inside or outside of the tank.

The Buchholz relay is usually used for transformers with ratings above1000 kVA and in some exceptional cases between 100 and 1000 kVA.For larger transformers, the Buchholz relay is used together with thedifferential relay as an important protection against faults. The first stageof the Buchholz relay is very sensitive and detects very small gasemissions, which cannot be detected by the differential relay respondingto higher measurable currents, e.g. above 20 % of the rated current. Onthe other hand, high fault currents are detected by the differential relayand by the Buchholz relay which both initiate a trip and interruption withinapproximately the same time.

In the US a Sudden-Pressure Relay (SPR) is usually used instead of theBuchholz relay. There are two types of SPR devices, operating on suddenchanges- in the gas above the oilor- in the oil itself.

Both types are sensitive to low-and high-energy arcs and have inverse-time characteristics. Generally the SPR devices are used to trip. They arefree from false operation.

3.6. Stator Ground Fault Protection

The stator ground fault protection protects the generator stator againstdamage caused by ground faults.

The consequence of a ground fault is the damage of the insulation andadditionally the damage of laminated iron of electrical machines. Thedamage depends on the level of the fault current; therefore this currentshould be kept very small. Higher ground fault currents on rotatingmachines can lead to short-circuits in the iron lamination and to higheriron losses.

The stator ground fault protection system is different for generatorsworking with a generator transformer or working directly to a busbar.

3.6.1. Stator Ground Fault Protection for Generators Operating inConnection with Generator Transformers

The following stator ground fault protection devices are applied forgenerators with an impedance-grounded or an ungrounded neutral point.For the detection of a stator ground fault either the generator voltage or anauxiliary voltage can be used. Using the generator voltage, approximately


31

95 % of the generator winding lengths can be protected; the rest of thewinding near the neutral point remains without protection.

An artificial voltage displacement of the generator neutral point by anauxiliary voltage is to be used if the winding at the neutral point shouldalso be protected.

3.6.1.1. 95% Stator Ground Fault Protection

The stator of generators operating in a unit with the transformer is usuallyhigh resistance grounded. The task of the grounding resistor is a sufficiantselectivity of the applied ground fault protection device. Stator groundfaults cause a voltage displacement of the generator neutral point. Thisdisplacement depends on the point of the ground fault. It is zero at aneutral point ground fault and it is equal to the generator phase voltage ata terminal ground fault (Fig. 3.6-1).

The voltage displacement of the neutral point can be measured either- between the generator neutral voltage and the ground (Fig. 3.6-1)or- by the voltage across the broken delta of the tertiary windings of three VT's located at generator terminal side (Fig. 3.6-2).

It is also possible to measure the current of the grounding resistor insteadof the neutral point voltage.

The neutral point voltage consists of several components, two of whichoccur only under fault conditions; the other two components occur alsoduring normal operating conditions. These voltage components are:

Va - Voltage caused by a ground fault and depending on its positionVb - Voltage transferred from the H.V. side at a ground fault on the H.V.

sideVc - Voltage of the 3rd, 9th, 15th, 21st and higher harmonicsVd - Voltage caused by asymmetry of phase capacitances and phase

voltages.

During a ground fault of the neutral point the component "a" is equal tozero. This voltage increases with the rising distance from the neutral pointand reaches the value of the full phase voltage at a fixed ground fault onthe generator terminals.The measured voltage may be higher than the phase voltage value duringan arcing ground fault due to oscillations produced between the statorcircuit and the ground. The neutral point voltage transformer should neverbecome saturated with respect to a danger of ferro-resonance. Thereforeit is to be designed for a higher voltage than the phase voltage. The bestmethod is to specify the VT for the voltage factor 1.9 (acc. to IECstandard) or for the line-to-line voltage (acc. to ANSI standard).

Also the three VT's of the variant with the VT's located at the generator


32

terminal side should be specified by the voltage factor 1.9 or by the line-to-line voltage value.

A modified method uses the current measurement instead of the voltagemeasurement. In this case a current transformer is connected betweenthe grounding resistor and the ground (refer to Fig. 3.6-3).

In case of a ground fault on the H.V. generator-transformer side, thephase voltage of the H.V. side produces a certain voltage transferred tothe generator side (component "b") by the capacitive coupling.

During normal as well as abnormal operating conditions the neutral pointvoltage contains the 3rd, 9th, 15th, 21st and higher harmonics of thecomponent "c" and a component "d" given by the asymmetry of the phasecapacitances and phase voltages.

Taking into account a danger of a protection maloperation the suitablesetting of the measured voltage must be selected with respect to thespurious components "b", "c" and "d". At an usual protection devicesensitivity of 5 % the total value of spurious components should be equalto or below 2.5 %. Based on such a sensitivity, 95 % of the winding lengthis protected, naturally under the assumption of a linear voltage distributionalong the winding.ABB devices are usually delayed for 0.5 s, in order to exclude anymaloperation at transient voltages.

Property of the Protection Method

ABB prefers the method using a grounding resistor and a voltage trans-former connected in parallel between the generator neutral point and theground for the following reasons.

In both cases (Fig. 3.6-1 and 3.6-2) the grounding resistor reduces thetransferred voltage of the component "b" and enables a selectivedetection of ground faults on the generator side. If this resistor fails or isinterrupted, the selectivity of the protection may be lost, but the generatorside remains protected.

The voltage transformer located at the neutral point is without any voltageunder normal operating conditions. It is energized only during a groundfault and is then slightly loaded by the very small power consumption ofthe protection device.

The three voltage transformers at the generator terminals operate with arated voltage under normal running conditions. During a ground fault theirvoltage is different for each phase and in the range between zero and theline-to-line value. For a short time the voltage transformers are overloadedby the current of the grounding resistor connected to the broken delta oftertiary windings.


33

The method of monitoring the current circulating between the generatorneutral point and the ground is simple; the sensitivity is the same as forthe voltage method. The detection is lost by an interruption of thegrounding resistor.

Grounding Resistor

The grounding resistor has to restrict the influence on the voltagedisplacement of the generator neutral point by:- voltages transferred from the transformer during H.V. ground faults- transient overvoltages produced by generator arcing ground faults and circuit breaker restrikes.

The above mentioned reduction of transferred voltages requires a resis-tance value which is as small as possible. Such a low resistance valueincreases the ground fault current. On the grounds of test results, ABBrecommends to limit the ground fault current to 20 A for a duration of 0.5s. Higher currents cause damages making the repair of the iron laminationnecessary. At ABB the grounding resistor is usually specified for currentsbetween 5 and 15 A during 10 seconds.

Dimensioning of Grounding Resistor

The grounding resistor is to be designed according to two conditions:

- Maximum ground fault current IE < admitted value IEmax < 20A- Spurious voltage Vs < 0.5 of the setting value Vset

Used symbols:

IE, IEmax ground fault current

Vs, Vb, Vc, Vd spurious voltages

Vset setting voltage value

V neutral point voltage

Vph generator phase voltage

RE grounding resistance

U2ph phase voltage of the H.V. generator transformer winding

C capacitance equivalent to three phases of the generatorside to ground; it consists of the following capacitances:

Co generator capacitance


34

Ce capacitance of the lead connection to the generator transformer

C1 capacitance of the L.V.winding of the generator transformer

L inductance equivalent to all voltage transformersconnected to the three-phase system of the generator.

C12 capacitance between the primary and secondarywinding of the generator transformer per phase

� angular velocity

The capacitance

C = 3 (Co + Ce + C1)

Based on the first condition of the maximum ground fault current, it isnecessary to calculate and specify the maximum current of the groundingresistor.

The second condition is used for the determination of the maximum valueof the grounding resistance with respect to spurious voltages. A spuriousvoltage consists of three components. The equation for these three com-ponents expressed in p.u. values:

vs = vb + vc + vd

The neutral point voltage displacement causes a ground fault currentwhich is limited by the impedance ZE of the generator stator side toground. The influence of the winding reactance is small and can beneglected. The impedance ZE has generally a capacitive component, acertain inductive component given by the VT's and an additional resistivecomponent given by the grounding resistor.

Using the total capacitance to ground C, the inductance L equivalent forall VT's applied at the generator terminal side (per three phases) and theresistance RE of the grounding resistor we get the following formula forthe impedance:

Z R

j CR LCLC

EE

E

�

��1 1

��

�

2

2

The value � 2LC has to be checked. If it is � 2LC >> 1 as usual, then


35

Z Rj CRE

E

E

�

�1 �

or

R Zj CZE

E

E

�

�1 �

We put ZV

IEph

E

�

max

and get e.g. for

V Vph �

240003

I AEmax � 10

C x F��0 6 10 6. (per 3 phases)

ZE �24000

3 10 = 1385 ohms

For the connection of RE and C in parallel:

� �1

Z1

RC

E

2

E

22�

��

�

��

�

��

�

��

� �R

Z

1 CZE2 E

2

E2�

� �

� �R

13851 314x0.6x10 x1385Emin

2

6 2�

��

for � = 314 1/s

REmin > 1435 ohms

The ABB static devices applied for monitoring the neutral point voltagedisplacement are less sensitive to higher harmonics and therefore

vc = 0

The componnent "d" is usually smaller than 1 or 2 % of the phase voltage.

Using the value vd = 2 % and keeping the value vb < 2.5 % a setting valuevset = 5 % is not exceeded by spurious voltages.

The transformer capacitance C12 is much smaller than the generatorcapacitance C0. Therefore for an approximate calculation of a transferredcurrent during a ground fault on the transformer H.V. side only thecapacitance C12 may be applied:


36

a) Transformer H.V. neutral point ungrounded

I 3 C VC 12 2ph� �

The voltage drop across the impedance ZE

�V Z IE C�

should be equal or smaller than 2.5 % of the generator phasevoltage Vph. Then

�V V Z Iph E C� �0 025.

and the maximum impdedance and resistance value is

ZV

IEph

C

�0 025.

� �R

Z

1 CZEmax2 E

2

E2�

� �

Using the values

V Vph � �

240003

13856

� � 314 1/ s

C x F1293 10�

�

V Vph2110000

363500� �

C x F��0 6 10 6.

we get

I x x x x AC � ��3 314 3 10 63500 0 189 .

Z x ohmsE � �

0 025 138560 18

1924..

Rx x xEmax ( . )

�

��

19241 314 0 6 10 1924

2

6 2


37

REmax < 2064 ohms

The final resistance value is to be chosen between REmin and REmax, it isbetween 1435 and 2064 ohms. Taking the value of e.g. 1500 ohms we getfor themaximum resistor current

IVR

ARph

E

� � �

138501500

9 23.

- maximum ground fault current

IVZE

ph

E

�

Z Rj CRE

E

E

�

�1 �

Zj x x xE �

��

15001 314 0 6 10 15006.

Z ohmsE � 1443

I AE � �

138501443

9 60.

b) Transformer H.V. neutral point solidly grounded.In this case only 1/6 of the value of 3 C12 is activ.

I C VC ph� 0 5 12 2. �

I x x x x AC � ��0 5 314 3 10 63500 0 039. .

Z x ohmsE � �

0 025 138560 03

11547..

Comparing this maximum impedance value ZE with the reactance value ofthe capacitance C

XC x

ohmsC � � �

1 10314 0 6

53086

� .

we see that the capacitance C allone is able to keep the voltage dropproduced by a current transferred from the H.V. side, below the level ofthe ground fault protection pick-up value. The resistor RE is thus notdesired to stabilize the protection device at H.V. ground faults; it is to be


38

used only to damp transient voltage oscillations. Its value should thus beequal or smaller than the capacitive reactance Xc.

The final resistance value can be chosen between 1435 and 5308 ohms.Taking the value of e.g. 3000 ohms we get:

I AR � �

138503000

4 62.

Zj x x xE �

��

30001 314 0 6 10 30006.

Z ohmsE � 2612

I AE � �

138502612

5 3.

For a resistor connected to the secondary side of a voltage transformersituated between the generator neutral point and the earth the resistancevalue Re must be recalculated:

R RVVe E

2N

1N

2

��

��

�

��

The voltages V1N and V2N are rated voltages of this voltage transformer.The value of a resistor connected to a tertiary side in delta of 3 voltagetransformers situated at generator terminals is:

R R3VVe E

3N

1N

2

��

��

�

��

The voltage V3N is the rated tertiary voltage of a VT.

If the transformer capacitances are not known, the following maximumdata can be used for an approximate calculation:

Capacitance of 1 phase (pF)C1 5000 - 12000C2 1000 - 1500C12 3000 - 6000

where:

C2 is the H.V. generator transformer winding capacitance( not important for the determination of the grounding resistor)


39

3.6.1.2. 100 % Stator Ground Fault Protection

The task of the 100 % stator ground fault protection is the detection of allstator ground faults.

The 100 % stator ground fault protection comprises a 95 % scheme andan injection scheme. The injection scheme operates with a neutral pointdisplacement by an injected voltage. At the REG216 and 316 systems theinjection voltage has an impulse waveform with an amplitude of about100V and a frequency of 12.5 or 15 Hz.

The protected winding length depends on the maximum current of thepower system frequency circulating through the grounding resistor REduring a terminal ground fault. If this current is smaller than 5 A, the totalwinding length of 100 % is protected by the injection scheme. If thiscurrent is higher, e.g. 10 A, then only 50 % of the winding length from theneutral point is protected by the injection scheme.

The advantage of the low frequency for the injection voltage is the:- possibility to distinguish between a service and a ground fault current- low capacitive currents during normal operation.

Typical settings:

95 % scheme Voltage 5 or 10 % Delay 0.5 s

Injection scheme Insulation resistance Delaystage No.1 "alarm" 10000 ohms 1 sstage No.2 "trip" 1000 ohms 0.5 s

Remark: a similar injection scheme can also be applied for the rotorground fault protection.

3.6.2. Stator Ground Fault Protection for Generators Directly Connectedto a Busbar

This ABB protection system is applicable for ungrounded systems(Fig. 3.6-6).

The purpose of this protection system is- on the bus and transformer feeders: to give an alarm- on generator feeders : to trip the faulted feeder during ground faults.

Any ground fault on the system is detected by a voltage displacementmeasured on the busbar, which is common for all machines of thissystem.


40

The voltage displacement is measured by a sensitive voltage deviceconnected to the broken delta of the tertiary windings of three voltagetransformers supplied by the busbar voltage. In order to selectively detectand remove the feeder affected by the ground fault, a current or better adirectional current measurement is necessary in addition.

For an ungrounded power system the ground fault currents aredetermined by the capacitances to ground. In most cases these currentsare small, not well measurable and cannot be used for the detection of aground fault. In these cases it is necessary to increase the level of theground faults by a grounding impedance. Usually a grounding resistor isapplied.

The ABB protection system uses a grounding resistor Re limiting theground fault current to 20 A. As mentioned, when a ground fault occursthe voltage device picks up. This device sounds an alarm. Consequentlythe voltage device initiates the connection of the grounding resistor to thebroken delta of 3 VT's (Fig. 3.6-7). A sufficient ground fault current cannow flow. According to the location of the ground fault, the ground faultcurrents vary between 0 and 20 A.

For a ground fault on the busbar and on the transformer feeders nocurrent device can pick up, and therefore the grounding resistor will bedisconnected after 3 - 6 s.

The grounding resistor and the grounding transformers are full-current-rated only for a short time operation of 10 s. The purpose of the resistorRp is to reduce the influence of spurious voltages for a ground fault on theH.V. side. The resistor Rp is dimensioned for continuous operatingconditions.

For a ground fault on the generator side of the current transformers thedirectional overcurrent device picks up and the faulted generator is trippedafter 0.5 s.

Features of this method:- very sensitive and complex- not available for large machines. In the case of large machines it is not possible to detect small ground fault currents below the relatively high rated current.- the special zero sequence current transformers are required- the grounding transformer is overloaded for a short time- the grounding resistor is dimensioned for a short time only- reference voltage value depends on the location of the ground fault.

Summary:- The voltage device detects any ground fault


41

- If only the voltage device picks- up, it means that the bus or some transformer feeder is faulted to ground- The directional current device detects ground faults on its generator feeder.

If a running generator is not yet connected to the busbars, a ground faulton its feeder cannot be detected. In this case it is possible to use anadditional neutral point voltage transformer (see Fig. 3.6-7) and a voltagedevice to detect ground faults on its feeder. A grounding resistor is notrequired, because there is no transferred current as a spuriouscomponent. This protection device trips the excitation circuit breaker.When the generator circuit breaker is closed, this additional protectionsystem is blocked.

3.7. Rotor Ground Fault Protection

The rotor ground fault protection is the protection of the field circuitagainst ground faults.

The rotor ground faults are less dangerous, because the fault current issmall due to the low voltage. More dangerous is the double rotor groundfault causing heavy rotor vibrations due to the magnetic unsymmetry.

For generators with sliprings the rotor insulation resistance is sometimesreduced by a coal dust layer produced by the coal-brushes.

The applied rotor ground fault protection device is not able to differentiatebetween single- and double-ground faults. Depending on the kind of theexcitation system, the ground faults of the excitation machine, rectifier andexcitation transformer are also detected.

The rotor circuit is ungrounded under normal operating conditions and theexcitation voltage is divided into two voltages of the + and -pole to ground.At the occurrence of a ground fault this voltage distribution is changed andresults in a small ground fault current.

Concerning the capabilities of the measuring technique, this small DCcurrent, compared with the much higher field current, is not suitable to beapplied for the ground fault detection. Instead a small AC voltage isinjected between the field circuit and the ground according to Fig. 3.7-1.This voltage value has to be limited to a level of 50 V for the safety of themaintenance personnel. The capacitor C blocks any DC current to avoidsaturation of the supplying transformer as well as of the measuringtransformer.

The simple rotor ground fault protection system monitoring the current ofthe supplying voltage transformer is suitable for generators with excitationsystems having a voltage with only few harmonics, such as e.g. excitation


42

systems with diode rectifiers. Larger harmonics would cause faultyoperation at this protection system.

At generators with modern excitation systems, using controlled thyristors,various harmonics as well as a certain fundamental frequency voltagecomponents can occur in the excitation voltage. For that application a newprotection device was developed. This device, using a balanced bridgeconnection (Fig. 3.7-1), is less sensitive to the above spurious com-ponents. During commissioning the measuring bridge circuit must bebalancend according to the rotor capacitance C by the capacitor C2.During a ground fault the capacitance C is shorted and the bridge isgetting unbalanced.

3.8. Transformer Ground Fault Protection

3.8.1. Typical Ground Fault Protection

Transformer ground faults are detected by a Buchholz relay or by asuitable protection device for windings connected to

- an ungrounded system: by supervision of the voltage across thebroken delta circuit of VT's (Fig. 3.8-1)

- an impedance grounded by supervision of the neutral point current system: (Fig. 3.6-3)- a solidly grounded system: by the differential protection

The protection device of an ungrounded system using the voltage super-vision across the broken delta is an unselective protection; the groundfaults on the winding as well as on the connected system produce such avoltage. During ground faults on the connected lines the measuredvoltage is 100 %; during ground faults on the transformer winding themeasured voltage depends on the faulted point. At a delta connectedwinding the minimum measured voltage is 50 %. A suitable setting valuewould be e.g. 40 % for the detection of local and remote ground faults. Forthe detection of remote ground faults on lines a setting value between 60and 80 % may be sufficient.

If the voltage level of the other transformer winding is higher, as is thecase with generator transformers, it is to be calculated with transferredcapacitive currents from the H.V. side and the circuit of the broken delta isto be closed by a resistor or by a nonlinear element.

Also the device operating with the neutral point current of an impedance-grounded system (Fig. 3.6-3) is an unselective protection detecting groundfaults not only on the transformer winding but also on the connectedsystem. The maximum ground fault current circulates during ground faultson the terminals or on the connected lines. Evaluating the maximum


43

ground fault current as 100 % and using a current setting of 20 %, theprotected winding length is 80 %.

On a solidly grounded system the level of ground fault currents is similarto the level of short-circuit currents. Therefore a differential device is ableto detect the currents of both kinds of faults.

3.8.2. Restricted Ground Fault Protection

The restricted ground fault (R.E.F.) protection is a selective protection(Fig. 3.8-2).

The R.E.F. protection is used especially for impedance-grounded trans-former windings protected by the transformer differential protection.Its task is to improve the ground fault protection extending the protectedWye-winding length. During ground faults on the wye connected windingthe ground fault current induces a current into the delta winding. Thisinduced current is smaller than the ground fault current (both in p.u.system) due to a changed turn ratio. The differential protection operateswith the smaller current i1 whilst the R.E.F. protection operates with thehigher current i2, which is more suitable for the detection.

E.g. the R.E.F. protection enables the protection of 80 % of the windinglength, whereby only 42 % can be covered by the differential protectionwith the same setting, when the maximum ground fault current is limited tothe value of the rated current by the grounding impedance and the settingis 20 % for both devices (Fig. 3.8-2).

For the detection of faults instantaneous voltage or current devices areused. The current device is used with a series-connected stabilizingresistor (Fig. 3.8-3).

The measuring system is usually a high-impedance system; its advantageis a very good stability for external faults.

For internal faults the fault current causes high overvoltages across thehigh-impedance of the differential circuit. These overvoltages must belimited by one or more nonlinear resistors connected in parallel to thedifferential circuit. Sometimes the differential circuit is to be shorted afterthe protection device has tripped.

Only current transformers with the same current ratio can be used. Theyare to be specified by the knee-point voltage and by the maximum allowedsecondary winding resistance.

Although the advantage of the R.E.F. protection is most important forwindings with impedance grounded neutral points, this protection is alsoused for solidly grounded windings.


44

The R.E.F. is to be designed in two steps:

step No.1 - determination of the R.E.F. device and of its settingaccording to an external ground fault or short-circuit

step No.2 - determination of the CT's acc. to an internal ground fault.

The determination of the 1st step is required to avoid maloperation due toCT saturation during through-ground-fault currents as well as externalshort-circuits.

3.8.3 Tank Leakage Protection

The tank leakage protection is a selective protection against ground faultsbetween the transformer winding and the tank. It can be applied only if thetank is fully insulated from ground and is connected to the ground by acurrent transformer supplying a ground fault current device (Fig. 3.8-4).

The ground fault current circulates from the tank to the ground during aninternal fault and is monitored by the current device.

The protected zone, generally limited to the tank, can be extended tocables if their sheath can be insulated from the ground in a similar way asthe transformer tank.

3.9. Breaker Failure Protection

If a fault in a protected zone cannot be cleared due to a C.B. failure, thetripping command must be extended to the next C.B. or CB's. This remotetripping must be fast enough to limit the damage caused by the fault. Thebreaker failure protection B.F. can be realized by two different schemesusing as criterions (Fig. 3.9-1):

a) a pick-up signal of an overcurrent device and trip command (device50/62 BF)

orb) a C.B. image and trip command (device 62 BF).

The BF protection can be arranged with one stage only to trip the remoteC.B.'s or with two stages. In the second case the first stage is provided torepeat the trip of the local C.B. and the second stage to trip the remoteC.B.'s.

3.9.1 B.F. Device 50/62 BF

The scheme with the overcurrent device and the trip command is mostlyused.


45

It uses a trip signal and a current still circulating and not yet interrupted asa criterion of a C.B. failure. At a false trip signal a maloperation of the B.F.failure device is prohibited with the help of the current supervision.

During correct operation of the C.B. the fault is cleared after a time givenby:- the operating time of the fault detecting device- the C.B. operating time.

After the fault has been cleared, the overcurrent device drops out after thereset time. If the overcurrent device does not drop out, this means that thecurrent still circulates due to a C.B. failure. In this case the fault currentmust be interrupted by a next C.B. or by next C.B's tripped by the C.B.failure protection after a delay set on the timer 62.

In order to keep a short operating time the reset time of the overcurrentdevice must be as short as possible.

The CT's applied to supply the current device of the B.F. protection shouldbe located close to the protected C.B. The B.F. protection is provided forthe detection of currents circulating trough the failed circuit breaker. If theapplied CT's are located too far from the C.B., only the current supplied bythe generator is monitored during faults between the CT's and the C.B.and can cause a B.F. device maloperation although the C.B. is alreadyopen (Fig. 3.9-2).

The minimum operating time of the device 50/62 BF is given by theoperating time of the failed C.B. and by the reset time of the overcurrentdevice at the standard version (Fig. 3.9-3). It can be e.g. 70 + 30 = 100ms. The setting value of the operating time should be higher by a certainmargin. The operating time is to be set on the timer 62. Together with theoperating time of the fault protection (e.g. of 30 ms) and with the operatingtime of the next C.B. (e.g.70 ms) the total fault clearing time would be 200ms.

The modified B.F. device has a minimum operating time given by the ope-rating time of the failed C.B. only.

The current setting must be selected such that the overcurrent devicedoes not drop out due to the damping of the fault current during theoperating time of the B.F. device.

A typical current setting is above the rated current value. Applying a valuebelow IN the range of the current supervision is extended down to theservice current values, but the risk of a maloperation is higher.

If the reset time of the overcurrent device is too long a modified B.F.


46

device can be used. Its overcurrent device 50 BF is normally blocked andis released only if the timer 62 is timed out.

A disadvantage of the device 50/62 BF is that it is not applicable duringfaults with no fault overcurrent such as- overvoltage, overexcitation, frequency- unbalanced load, ground faults on ungrounded systems- loss-of-excitation, pole- slipping, reverse power and C.B. failures at a normal operating current.

3.9.2 B.F. Device 62 BF

The protection scheme uses a trip signal and an auxiliary contact of theC.B. for the detection of a C.B. failure (Fig. 3.9-1,4).

Advantage: - The device is independent of the current level. - The device can also be used at unbalanced load,

loss-of-excitation, overvoltage, etc.

Disadvantage: - Increased risk of a maloperation by a false trip signal during normal operation- Maloperation during a fault and wrong operation of the C.B auxiliary contact.


47

4. PROTECTION DEVICES FOR ABNORMAL OPERATING CONDITIONS

4.1. Overload Protection

4.1.1 Introduction

The overload protection is a protection of the winding insulation againstoverheating which could influence the insulation life.

Electrical machines such as generators, motors and transformers aredesigned for a certain maximum ambient air temperature; mostly 40°C. Atload the winding temperature rises due to machine losses, especially tocopper winding losses. The winding and its cooling system are usuallydesigned such that the temperature rise is 60 K for oil transformers and80 K for air cooled generator stators at rated current. This implies anabsolute continuous winding temperature of 100°C or 120°C. Thepermitted maximum continuous temperature value depends on theinsulation class and is usually higher by 10°C; that means an absolutetemperature of 110°C or 130°C. Higher winding temperatures cause ashorter insulation life and are therefore allowed for faults for a short timeonly. During normal service overloads, the maximum temperature value of110 or 130°C must not be exceeded.

For transformers with high through-fault currents a maximum short timecopper winding temperature of 250°C is allowed.

The insulation overheating can be caused by:- increased winding current- failure of the cooling system.

Various monitoring and protection devices supervise:- cooling gas or oil temperature- winding temperature rise- winding temperature.

The following components are used for overload protection:- Thermal sensors- Thermal devices.

4.1.1.1. Thermal Sensors

The spot temperature the of insulation surface, the cooling gas or the oil ismonitored.

Advantage: Critical points are able to be supervised

Disadvantage:- Only the insulation temperature instead of the windingtemperature can be measured, due to the high voltage


48

level.- Delay of the temperature detection

4.1.1.2. Thermal Devices

Thermal devices are provided for the supervision of the average tem-perature rise or of the average winding temperature. Because the directmeasurement of the winding temperature for high-voltage machines is notpossible, a thermal image of the winding temperature is used.

The thermal image usually operates with the assumption that thetemperature rise depends only on the winding current. If the currentsuddenly changes, the winding temperature rise follows according toseveral exponential functions for the temperature rise of the- cooling medium against the ambient air- winding against the cooling medium.

The transient functions of the variable temperature rise are specified bythe:- end temperature rise at a certain current step- thermal time constant.

E.g. a transformer oil temperature rise curve can be specified by the:- temperature rise of 55 K at IN- thermal time constant value between 120 and 210 min.A winding temperature rise curve can be specified by the:- temperature rise of 5 K at IN- thermal time constant value between 5 and 20 min.

The thermal image enables the simulation of the instantaneoustemperature rise without direct temperature measurement and without anyadditional delay.

The supervision of the winding temperature rise ensues independently ofthe ambient air temperature and the cooling system.

The winding temperature including the ambient air temperature andcooling intensity can be supervised using a thermal sensor in addition tothe thermal image.

Using two exponentional functions the variable winding temperature of anoil cooled transformer loaded by an overcurrent i can be calculated asfollows: (Fig. 4.1-1).

� � � � ��

w Noil

t

Nw Noil

t

t e e ioil w( ) ( ) ( )( ) ( )� � � � � ��

��

�

�

��

021 1 1

Equation 1.1


49

where i IIN

�

Used symbols:

�w t( ) - variable winding temperature as a function of time

t - time

�0 - ambient air temperature equal to cold machine temperature

��Noil - temperature rise of the oil at the rated current IN

��Nw - temperature rise of the winding at the rated current IN

� �oil w, - thermal time constants of the oil and winding

I , IN - instantaneous and rated current

Monitoring the instantaneous oil temperature �oil by a thermal sensor

� � ��

oil Noil

t

e oil� � ��

��

�

�

�

0 1 Equation 1.2

we get a simplified equation for the winding temperature

� � � ��

w oil Nw Noil

t

t e iw( ) ( ) ( )� � � ��

��

�

� �

�

1 12 Equation 1.3

�oil - instantaneous oil temperature

For a gas or water cooled generator or motor similar equations are valid.

Although under transient current conditions the winding temperature chan-ges mainly in accordance with two exponential functions, most of thethermal devices operate with only one exponential function. This functionmust be as equivalent as possible to both of the above functions (refer toFig. 4.1-2). This equivalent exponential curve corresponds to the windingtemperature rise and has a certain time constant � with a value betweenthat of �oil and �w . The thermal image thus operates according to theequation:

� ��

w Nw

t

e i� ��

��

�

� �

�

1 12( ) Equation 1.4


50

For an optimum value of " � " refer to Fig. 4.1-2. Often a suitable value is� = 0.7 �oil .

4.1.1.3. Equations for Operation Curve of Thermal Devices

According to the equation for an instantaneous winding temperature rise�� a certain temperature rise is reached at various currents duringdifferent times. We can consider an allowed maximum operatingtemperature rise, e.g. of 70 K, as a such temperature rise. The locus of allpoints with the allowed maximum temperature rise is a curve of thethermal capability of the protected machine (Fig. 4.1-4). It is a pattern forthe characteristic of the applied thermal device which has to operate atthe same or shorter times.

Used symbols:

IN rated current

I current at overload

Im current at maximum temperature

�0 temperature at I = 0

�N steady-state temperature at IN

� steady-state temperature at I

� i instantaneous temperature at time ti

�m maximum temperature

ti instantaneous time

tm overload duration until �M is exceeded

� thermal time constant

K multiplying factor

The end value of the temperature rise is given by the current (Fig. 4.1-3):

�� 02KI Equation 1.5

�� N N NKI� � �02 Equation 1.6


51

�

�

�

�N N

II

�

2

2

The instantaneous temperature:

� � � � �

i N N

t

ei

� � � ��

��

�

�

�

( ) 1

or

� � � � �

1 1� � � ��

��

�

�

�

N N

t

ei

( ) Equation 1.7

� � � �

1

2

1 1� ��

��

�

��

��

��

�

��

�

��

�

N NN

tII

ei

� Equation 1.8

The device picks-up at

� �i m� ti = tm

� � � �

m N NN

tII

em

� ��

��

�

��

��

�

��

��

��

�

��

�

�

2

1 1 Equation 1.9

�� m m N� �

According to the last and following equations the allowed maximum tem-perature rise ��m can be reached at various currents I during differenttimes tm:

� �

�

�m N

N

N

t

II

em�

�

��

�

��

� ��

1

1

12

� � ��

�

��

�

��

��

�

��

tn e n

II

m m N

N

N

�

� �

�� 1

1

12�


52

t n

II

mm N

N

N

�

��

�

��

�

��

��

�

��

��

�

�1

11

12�

Equation 1.10

The limits of the last equation are given by:

IIN

� �: tm � �� ln1 0

I Im� : tm � �

ln1

11

II

1

m N

N

N

2��

�

��

�

��

�

�

�

� ��

��

IIm

N

m N

N

� ��1 � �

��

For the overload capability of the machine according to the equation 1.10refer to Fig. 4.1-4.

The thermal time constant is

��

�hG

Sc

[s; J/kg; W/cm2; cm2]

Equation 1.11

where:

h specific heat [ J/kg ]G weight [ kg ]�c cooling factor [ W/cm2 ]

S cooling surface [ cm2 ]


53

4.1.2. Overload Device for Generators

4.1.2.1. Overload Device Characteristic

The tripping characteristic of this ABB device consists of two sections. Forsmall overcurrents a definite time curve and for high overcurrents aninverse time characteristic is provided. The characteristic of the secondsection is specified according to the overload requirement by theAmerican Standard ANSI C 50.13 for turbogenerators. The appliedcharacteristic is given by an approximate equation, which is suitable atfast short time current changes. Using a series development of theexponential function we obtain:

e x x xx�� 1 1

112

13

2 3

! ! !...

1 11

12

13

2 3� � � � �

�e x x xx

! ! !...

Usig the equation 1.8 we get

� � ��

m N NN

i i iII

t t t� �

�

��

�

��

��

�

��

��

��

��

��

�

�

��

2 2 3

112

16

...

Approximately for ti � 0 1. � it can be written:

� � ��

m N NN

iII

t� �

�

��

�

��

��

�

��

�

2

1

The tripping time tm is reached when

� �i m� t ti m�

tII

mm N

N

N

��

�

��

�

��

��

�

1

12

This approximate equation is correct for small ratios ti

�

�� 1; it results in

shorter tripping times for overcurrents of longer duration than the exactsolution.

For a characteristic according to ANSI for turbogenerator stators thefollowing is valid:

��

�

m N

N

��

�41 4.


54

The bottom of the definite time curve for stators is at 1.16 In and 120 s(Fig. 4.1-5). A similar characteristic can be applied for turbogeneratorrotors.

4.1.2.2. Thermal Device for Universal Heating Curve

If the heating curve of the protected machine cannot be represented by asuitable exponential function a universal heating curve can be used forthis thermal device of the REG system. The heating curve is to be devidedinto 40 steps. By a temporal derivation of these steps an impulseresponce function is obtained, desired for setting values. For details referto the Instructions 1MDU02005.

4.1.3. Requirements of Standards

The maximum permissible continuous winding temperatures ( IEC 34-1,VDE 0530 ) based on the insulation class B and on the ambient tempe-rature of 40°C are as follows.

Generator stator air cooled winding:

temperature rise 80 Kabsolute temperature 120°C

Transformer oil cooled winding

oil temperature rise 60 Kwinding temperature 105°C

Short-time overload capability at full load conditions

Generator IEC ANSI

stator winding ti

�

�

37 512

. ti

�

�

41 412

.

rotor winding - ti

�

�

31 812

.

According to the IEC / VDE standards the stator of generators up to1200 MVA may be overloaded by 1.5 IN during 30 s.This value cor-responds with the above formula of IEC.

Transformers:

maximum permissible copper windingtemperature at through-fault currents: 250°C


55

In order not to exceed the value of 250°C the duration of the through faultcurrent is to be limited to 2 s according to VDE and ANSI / IEEE.

4.1.4. Thermal Device Setting

The setting depends on the winding insulation class and on the windingtemperature at full load.

E.g. oil cooled transformers are often designed for the following tem-peratures at rated current and at an ambient temperature of 40°C:

oil 95°Cwinding 100°C

The corresponding temperature rise is:

oil 55 Kwinding 60 K

In view of the allowed max. winding temperature of 105°C or 110°C,possible settings of the thermal devices supervising the windingtemperature rise, would be as follows:

single stage devices 65 ... 70 Kdouble stage devices

1st stage (alarm) 65 ... 70 K2nd stage (trip) 70 ... 75 K

The recommended settings for devices with the scale in percent:

single stage devices 110 .. 120 %double stage devices

1st stage (alarm) 110 .. 120 %2nd stage (trip) 115 .. 125 %

Possible settings of thermal devices supervising the absolute winding tem-perature with the help of thermal sensors are:

single stage devices 105 .. 110°Cdouble stage devices

1st stage (alarm) 105 .. 110°C2nd stage (trip) 110 .. 115°C

Similar values can be determined for generators and motors.


56

4.1.5. Thermal Time Constants

Generally the values of thermal time constants for generators and trans-formers vary in a wide range depending on the machine design. Also thevalues of the maximum temperature of the cooling medium and of thewinding can be different depending on the insulation class. Therefore it isrecommended to get those values from the machine manufacturer.

For oil transformers the range of usual thermal time constants is asfollows:

�w 5 ... 10 ... 20 min�oil 120 ... 210 min

If the values of the individual thermal time constants are not known, theapproximate equivalent value can be determined from the measuredwinding temperature rise curve. For this purpose the low-voltagetransformer side is usually shorted and the high-voltage side is suppliedby a voltage reduced approximately to the value of the impedancevoltage.

At certain time intervals the transformer supply is interrupted for a shorttime for the winding resistance measurement. The winding temperature isthen calculated from the resistance rise.

If the transmission lines between the circuit breaker and the transformershould also be protected, then it may be necessary to take their thermaltime constants into account.

Generally these are within the following range:overhead lines 5 ..... 30 mincables 15 ....100 min

4.2. Negative Sequence Current Protection

4.2.1. NPS Current Protection for Generators

The negative sequence current protection is a protection of the syn-chronous machine rotor against dangerous overheating caused by un-balanced stator currents. The negative sequence current devicemonitoring stator currents is often called the NPS current device; thatmeans a negative phase sequence device.

The task of the NPS protection is chiefly the protection againstunbalanced load. If possible this protection should also operate duringasymmetrical faults not cleared by the fault protection in time.


57

The accuracy of the NPS current detection must be sufficient to avoid anymaloperation at symmetrical three-phase fault currents.

A negative sequence current component I2 circulating during a certaintime causes a rotor temperature rise according to a thermal function fromthe initial value at rated conditions �0 to a new end value � . Usually asuitable exponential function is considered as equivalent to the actualtransient thermal function (Fig. 4.2-1). The allowed maximum temperaturerise must not be exceeded at:- continuous current I2�- any short-time current I2.during time tm

The NPS current protection operates when the value ��m is exceeded.

4.2.1.1. Dangerous Conditions for Synchronous Machines

For synchronous machines any kind of operation with asymmetrical statorcurrents can be dangerous. This abnormal operation can be consideredas an operation with additional current components which increase themachine losses. These additional stator current components increase los-ses and temperature of the stator winding and especially of the rotorsurface temperature by induced eddy currents.

Asymmetrical stator currents can be split up in three symmetrical com-ponents, the positive, negative and zero-sequence components.

The negative sequence current component produces a magnetic field rot-ating in the opposite direction to the rotor with synchronous speed. In therotor there are induced currents doubling the rated frequency.

The zero sequence current component causes a magnetic field with afixed axis and a variable amplitude. The currents induced in the rotor havethe rated frequency.

Most generators are high-resistance grounded and the zero-sequencecurrent can flow to the ground only during a ground fault. In the ABBgenerators the ground fault current is limited to 20 A so that it is lessdangerous for the machine. Nevertheless 0.5 s after any stator groundfault occurs the generator is tripped.

Another source of a zero sequence current is a sudden line-to-generatorneutral point short circuit. In this case the zero-sequence component ofthe generator current is very high, and this fault must be quickly cleared bya fault protection. The probability of a such fault is very small at modernconstruction arangements. Therefore it is expected that such faults aredetected by the differential protection and no special zero-sequencecurrent protection device is desired.


58

The negative sequence current results in an increased rotor temperature.This is dangerous for the rotor surface and especially for some criticalspots of the rotor.Therefore a supervision of the negative-sequencecurrent is necessary; the NPS current device is to be applied, becauseno other device is able to supervise the negative-sequence current.

4.2.1.2. Unbalanced Stator Current Conditions

A negative sequence current occurs due to:- unbalanced load- asymmetrical faults.

An unbalanced load can be caused by:- single-phase loads of the supplied power system (railways and furnaces)- asymmetrical three-phase loads- circuit breaker failures.

The negative-sequence current is mostly a small and almost continuouscurrent. Nevertheless sometimes the negative-sequence current is higherand its changes are faster.

The negative-sequence currents circulating during asymmetrical faults arehigh transient currents (refer to Table 2-VI). Those fault currents must bequickly interrupted by a fault protection, so that the rotor temperature riseis kept less dangerous. During circulating fault currents, the negative-sequence current protection may pick up but it does not trip. Only a failureof the fault protection might induce the negative-sequence current deviceto trip.

Opening a circuit breaker under load or fault conditions can resolute in anasymmetrical current if the current is not interrupted by all three poles.Under load conditions the asymmetrical current caused by a breakerfailure is smaller than the rated current and cannot be detected by anyfault protection. The negative-sequence component of this current is verydangerous for the rotor (Table 2-I). In this case a breaker failure deviceinitiated by the NPS current protection device has to operate and trip thenext circuit breakers.


ABB use two different NPS current devices. They operate either with adefinite-time characteristic or with an inverse-time characteristic.

The task of the definite-time NPS current device is the protection atunabalanced load. The NPS current definite-time device trips with acertain delay in order to avoid any trip at short-time NPS currents caused


59

by sudden unablanced load changes or by remote faults before they arecleared by the fault protection. This device is not suitable to protect thegenerator during remote faults.

The task of the inverse-time NPS device is the protection during un-balanced load as well as during asymmetrical faults.

The negative phase sequence current devices (NPS current devices) eva-luate the current I2 from the measurement of individual phase currents.For selection of setting values the thermal rotor capability must be known,specified by the machine manufacturer with the help of- permitted continuous negative-sequence current i2 � (p.u.)- permitted product (i2)2 t (p.u.;s)

The permitted negative sequence current value i2 � for generators isusually in the range between 5 and 10 %.

The allowed product value (i2)2 t is usualy in the range between 5 and30 s.

For details of permissible values for machines of different types andvarious ratings refer to Tables 4.2-II and III.

With respect to low setting values of i2 at large machines, only currenttransformers with a sufficient accuracy are to be used.

ABB use the NPS definite-time devices in two stages for an alarm and atrip (Fig. 4.2-2). The current setting of both stages is selected according tothe value i2 ; the delay of the alarm stage is usually approximately one halfof the trip stage delay. The alarm delay should avoid any false alarm atshort-time unbalanced currents. The trip delay should avoid any trip duringlocal and remote faults before they are cleared by a fault protection.

The inverse-time device is preferred for higher short-time NPS currentsunder transient conditions caused by sudden unbalanced load changesand faults. This device operates according to the setting value of theproduct (i2)2 t. It enables a better protection at fast unbalanced loadchanges and at asymmetrical faults, because its setting can be selectedaccording to the rotor thermal capacity.

4.2.1.4. Application of NPS Current Definite-Time Device

The NPS device with a definite time characteristic usually has to protectthe generator only at unbalanced load. The expected values of the currenti2 are between 0 and 57.7 %. The typical current setting is 0.1 IN formedium rating machines. The delay for the trip stage must be

- longer than the operating time of all fast fault protection devices detecting the same current level; e.g.


60

t > 1 s

- shorter than the allowed time tm. This time is given by the allowed product

(i2)2 t and by the expected maximum current i2 = 0.577 IN .E.g. for (i2)2 t = 10 s the delay is

� ��

ti ti

100.577

30sm2

2

22 2� � �

A typical setting value for the alarm and the trip is 5 and 10 s.

4.2.1.5. Application of NPS Current Inverse-Time Device

The NPS inverse-time device operates according to a characteristic givenby the product (i2)2 t and by the current i2�. The REG 216 and 316 sys-tems enable the application of this device according to the requirementsof the IEC 34-1, VDE 0530 and ANSI C50.13 standards.

Used symbols for the calculation:

�O inital temperature

� end temperature I2

� i instantaneous temperature

�m steady-state temperature at I2

I2� maximum steady-state admissible negative sequence current

I2 instanteneous negative sequence current

IN rated current

� thermal time constant

K,K1, K2 multiplication factors

ti time at instantaneous temperature � i

tm maximum time as well as operating time

tmin, tmax times limiting the activated range of the inverse-timecharacteristic


61

Similarly as for the temperature rise caused by a symmetrial overload itcan be written for stator unbalanced currents:

�� O KI22 Equation 2.1

�� m m O KI� � ��2

2 Equation 2.2

� � �

i O

t

KI ei

� � ��

��

�

�

�

22 1 Equation 2.3

According to the condition that the same maximum temperature �m ispermitted for short time unbalanced loading as well as for steady-state un-balanced loading, we calculate the time tm, at which the temperature isexceeded:

ti = tm and � �i m�

� � �

m o 22

t

KI 1 em

� � ��

��

�

�

Equation 2.4

and

� �m O KI� ��2

2 Equation 2.5

Comparing the last equations we get

I 1 e I22

t

22

m

��

��

�

��

�

�

�

e

II

II

1

t2

2

2

2

2

2

m�

�

�

�

�

��

�

��

�

��

�

��

�

and later

t

II

II

m �

�

��

�

��

�

��

�

��

�

�

� ln

2

2

2

2

2

2

1

Equation 2.6

An approximate calculation is possible using the series development ofthe exponential function to the equation 2.3:

� ��

ii i iKIt t t

� � ��

��

��

��

��

��0 2

22 31

216

...


62

For small ratios ti�

an approximate solution using only the two first terms of

the series can be used:

� ��

ii iKIt t

� � ��

��

��

��0 2

221

2Equation 2.7

For the maximum permitted temperature

� �m i� tm = ti

we get

� ��

mm mKIt t

� � ��

��

��

��0 2

221

2Equation 2.8

� �m KI� ��0 2

2 Equation 2.5

Now we compare:

KIt t

KIm m22

2

221

2� ��

�

��

�

�

��

��

�Equation 2.9

I I t tm m22

22 1

112

�

��

��

�

��

�

� �

Equation 2.10

For tm

�

<< 1 can be written approximately

I It

Im

22

22

221

2� �

� �

� Equation 2.11

The operating time is

t I

I Im �

�

�

�

22

22

221

2

� Equation 2.12

or

t i

i i

Ki Km �

�

�

�

�

�

22

22

22

1

22

221

2

� Equation 2.13


63

where

i IIN

22

� i IIN

22

�

�

�

K i1 22

�� K i2 2

12

��

Equation 2.14

The REG system device "NPS-Invers" of ABB operates with curves givenby the equation 2.13; for one of its curves refer to Fig. 4.2-3.

The factor K1 is to be set according to the permitted product value of(i2)2t. Without the limitation by the "tmax" the operating time tm would by

infinitely long at the current i2 = K2 = 12 2i �

(refer to Equation 2.13).

The time "tmax" is the maximum permitted duration of the circulatingcurrent i2� This time is to be calculated from the permitted product value(i2)2 t:

tmax = � �i ti2

2

22�

At usual values for medium rating machines it is:

(i2)2 t = 10 s

i2 = 0.1

tmax = � �

100 1 2,

= 1000 s

The time "tmin" must be:

- longer than the operating time of all fast fault protection devices detecting the same current level; e.g.

tmin > 1 s

- shorter than the allowed time tm .

If the device has to protect the generator only at unbalanced load, themaximum value of i2 which should be considered is 0.577. From the value(i2)2 t = 10 the time tmin is


64

� �t smin .

� �10

0577302

A possible setting with respect to both conditions is e.g. tmin = 10 s. If thedevice has to protect the generator not only at unbalanced load but alsoduring asymmetrical faults, then the maximum value of i2, to beconsidered, is e.g. 3.33 during a generator line-to-line fault (refer toTable III). For the value (i2)2 t = 10 s the time tmin is

� �t smin .

.� �10333

0 902

The possible setting which can be realized with respect to both conditionsis

tmin = 1 s (minimum possible setting value)

At high currents I2 the second term tm

2� in the bracket of the equation

2.10 is quite small and may be neglected because tm

21

�

�� ; then we get a

second approximate solution:

I Itm

22

22

��

� Equation 2.15

� � � �t

II

ii

Kim � � �

� �2

2

22

2

2

22

1

22

� �

Equation 2.16

This second solution presupposes no heat dissipation and we get a verysimple formula for the rotor thermal capability

� � � �i t im2

2

2

2�

�� Equation 2.17

Some NPS current devices (not involved in the REG216/316 system) ope-rate according to this simple equation.

The value of the thermal time constant � depends on the rotor surfaceand is much larger than the thermal time constant of the winding.For instance at

K1 = 30 s , i2 = 0.1

the thermal time constant is


65

� ��

�

K

is or1

2

2

300 01

3000 50.

min

4.2.1.6. Level of Symmetrical Components at Various Conditions

The symmetrical components of a current system of the phases A,B,Ccan be calculated according to the following formula:

I1 = 13

( IA + a IB + a2IC )

I2 = 13

( IA + a2IB + a IC )

I0 = 13

( IA + IB + IC )

where "a" denotes a phase shift of 120°. The calculation of symmetricalcomponents for various conditions follows.

1. Symmetrical Current System

Referring to Fig. 4.2-4 we get

IB = a2 IA IC = a IA

and

I1 = 13

( IA + a3 IA + a3 IA ) = IA

I2 = 13

( IA + a4 IA + a2 IA ) = 0

I0 = 13

( IA + a2 IA + a IA) = 0

because1 + a + a2 = 0

a3 = 1 a4 = a

2. Interruption of 1 Phase

An interruption of e.g. phase B is followed by a change of current phasesaccording to Fig. 4.2-5. At an ungrounded system it can be written

IA + IB + IC = 0


66

For IB = 0 : IC = -IA

and the symmetrical components are

I1 = 13

(IA + 0 - a2 IA)

I2 = 13

(IA + 0 - a IA)

I0 = 13

(IA + 0 - IA)

Because 1 32� �a and 1 3� �a

we get

I1 = 13

IA

I2 = 13

IA

I0 = 0

The negative-sequence current is therefore 0.577 IA or 57.7 % of IA.Thismeans that the negative-sequence component reaches 57.7 % of thephase current at the full unbalance of phase currents caused by the load.

3. Reversed Phase Sequence

The supply of a symmetrical load by a negative-sequence voltage causesa reversed phase current sequence.

According to Fig. 4.2-6 we get

IB = a IA IC = a2 IA

and

I1 = 13

(IA + a2 IA + a4 IA) = 0

I2 = 13

(IA + a3 IA + a3 IA) = IA

I0 = 13

(IA + a IA + a2 IA) = 0


67

This shows that a 100 % negative-sequence current can be obtained onlyby a negative sequence voltage.

4. H.V. Line-to-Line Fault

Referring to Fig. 2-10

IA + IB + IC = 0

IB = - IA IC = - IA

and

I1 = 13

(IA - a 12

IA - a2 12

IA) = 12

IA

I2 = 13

(IA - a2 12

IA - a 12

IA) = 12

IA

I0 = 13

(IA - 12

IA - 12

IA) = 0

5. H.V. Line-to-Ground Fault

Referring to Fig. 2-10

IB = 0 IC = - IA

and

I1 = 13

(IA - a2 IA) = 13

IA

I2 = 13

(IA - a IA) = 13

IA

I0 = 13

(IA - IA) = 0

6. Testing by Single-Phase Supply

IB = IC = 0

I1 = 13

IA I2 13

=IA I0 = 13

IA

With respect to the equation for I2 the pick-up value is three times higherthan the setting value.

4.2.2. NPS Protection for Motors

The NPS current protection is also required for induction motors. The


68

negative sequence current causes a braking torque producing highercurrents and an increased rotor temperature. At motors the currentunbalance can be caused by:- Network voltage asymmetry- Circuit breaker failures- Motor faults.

The characteristic of the NPS motor protection device may be again adefinite-time one or an inverse-time one.

4.2.3. Checking of the NPS Current Device

Sometimes a test by an injected current is required.

At standstill of a generator it is possible to supply only one phase input ofthe applied device by an injected current. The pick-up value is then three-times higher than the setting value.

For the test a generator operating into a symmetrical short circuit arrangedby a link or by a grounding isolator can also be used. After shorting 1 or 2current inputs the NPS current device operates, if the current value issufficient. Never should an asymmetrical short-circuit of the generator beused for the tests because the danger for the rotor would be too high.

4.3. Voltage Protection

4.3.1. Overvoltage Protection

The overvoltage protection is the protection of electrical machines andcurrent circuits against damage caused by overvoltage and overheating.Particularly protected are:- windings of electrical machines and current circuits against dangerous electrical stress of the insulation- laminated iron of electrical machines against overheating caused by increased iron losses- stator conductors situated near the air gap of large machines against overheating from eddy currents due to the high intensity of the radial magnetic field.

The overvoltage protection is considered as a back-up protection for theA.V.R., which normally keeps the generator voltage within a certain toler-ance range.

The overvoltage protection is normally provided with two tripping stages.The first stage is intended against smaller and longer lasting overvoltages.The second stage is foreseen against higher and short-time overvoltages.Both stages have a definite time operating characteristic. A delay is


69

necessary to avoid any maloperation at short- time overvoltages causedby:- sudden load changes- asymmetrical short-circuits- switching overvoltages.

The generators are designed for a continuous operation at a rated voltagewith a tolerance of +/- 5 %, sometimes of +/- 7.5 % or +/- 10 %. Withrespect to a voltage device accuracy of +/- 5 % the typical setting valueof the1st stage is higher by 5 % than the highest continuous service voltage(Fig. 4.3-1):

1st stage: 1.10 VN for VN +/- 5 %1.13 VN for VN +/- 7.5 % delay 5 s1.15 VN for VN +/- 10 %

2nd stage: 1.30 VN delay 0.1 s

4.3.2. Undervoltage Protection

An undervoltage is no danger for the generator itself. Therefore an under-voltage protection is not normally applied for generators. An exception isthe case of a power system with an unusually low short-circuit currentlevel. An undervoltage caused by underexcitation could result in aninstability of the power system. In this case an undervoltage protection isto be used additionally to the "loss-of-excitation" protection.

The undervoltage protection is important for motors because the torquesinks with the voltage squared. Due to the reduced torque the motordriving capability and especially starting-up is made difficult. Therefore theundervoltage protection has usually two stages. The first stage is appliedfor blocking the motor energization and the second stage for tripping.

4.4. Overexcitation Protection

The overexcitation protection is applied for generators and transformersagainst overheating caused by the increased intensity of the magneticfield. This protection is especially required for modern transformers.An overexcitation can occur e.g. :- at rated frequency and overvoltage- at rated voltage and underfrequency.

An overexcitation occurs e.g. at start-up or running down due to a failureof the control system.

Overheating due to higher magnetic intensity in electrical machines is


70

caused by increased iron losses and by increased additional losses fromeddy currents.

Overexcitation of transformers is not only dangerous due to increased ironlosses and increased eddy currents, it increases also the magnetizing cur-rent to a value which could cause maloperation of the differential device.

The measuring quantity of the overexcitation device is the ratio of V/fwhich corresponds with the magnetic flux (at a neglected voltage drop onthe leakage reactance).

A short delay is recommended to avoid maloperation at sudden loadchanges such as load shedding, etc.

As a reference for setting values the diagram specified for Westinghousemachines (Fig. 4.4-1), the ANSI / IEEE Standard C37.106 or the VDEStandard 05320 may be applied.

For a protection device operating with a definite-time characteristic thetypical settings are:

1.10 VN/fN 5 s

This characteristic is suitable for a longer overexcitation. For a short-timeoverexcitation an inverse-time characteristic enables a betterprotection.The short-time overexcitation capability is different for variousmachines. For settings of a device using an inverse-time characteristic therequired machine capability data must be supplied by the machinemanufacturer.

4.5. Frequency Protection

The frequency protection protects rotating machines against vibration andoverheating.

Frequencies deviating below or above the rated frequency can causedangerous vibration of the generator and its driving machine. A consider-ably increased frequency can be dangerous due to large centrifugalforces. In this case the frequency protection operates as a back-upprotection for the speed regulator.

A higher frequency causes overheating of the magnetic circuits due to in-creased iron losses.

When selecting the settings the allowed continuous frequency tolerance of+/- 2 % for generators is to be considered. For allowed values of theprime-mover its manufacturer must be consulted.


71

4.6. Loss-of-Excitation Protection

The loss-of-excitation protection is applied to protect synchronousmachines, generators and motors operating in parallel with generators ofthe power system against dangerous conditions caused by under-excitation.

The field current of synchronous machines connected to a power systemmust be of a level which enables the supply of a required value of MWand MVAR to the power system (as a generator) or to be supplied by arequired value of MW and MVAR ( as a motor). If the field current is toolow the required value of MVAR cannot be supplied, the inductive currentof the stator is too low and could even get capacitive. Although adecreased field current cannot cause any stator or rotor overload atvalues of xd above100 % the temperature of stator end plates, of stator-end laminated iron and sometimes also of stator winding ends rises abovethe value during rated conditions. This is caused by an increased localmagnetic flux leakage. Depending on the short-circuit current level of thepower system its voltage could be decreased due to the underexcitation.The operation of an underexcited machine can result in instability and inthe loss of synchronism. If the underexcited machine decreases thevoltage of the power system, the instability of other machines may beinitiated.

An underexcitation can be caused by:- wrong setting of the A.V.R.- failure of the A.V.R.- voltage rise of the power system.

Dangerous conditions for an underexcited synchronous machine are asfollows:

Synchronous running with underexcitation:- Increased heating of stator end plates, stator end iron lamination and stator winding ends- Instability.

Asynchronous running after loss of synchronism:- Rotor surface overheating- Increased rotor mechanical stress- Field winding overvoltages (only at excitation systems with semiconductors).

With respect to the more dangerous consequences of "loss-of-sychronism" the protection is provided specially for this case. The usuallycalled "loss-of-excitation protection" is designed and set for the detectionof the majority of the service points at which the synchronism is getting


72

lost. The protection device operates by monitoring the angle between twophasors V and E being the load angle � (Fig. 4.6-3).

The conditions of an underexcited machine during synchronous operationare monitored by the A.V.R. A monitoring of the field current by an under-current device would not be efficient, because the field current variesbetween 100 and 250 % or more during operation.

A decreased field current of a generator operating in isolation causes adecreased voltage. The power factor is given by the load and does notdepend on the field current. An "underexcitation" such as known formachines operating in parallel with other synchronous machines of thepower system is not possible and the "loss-of-excitation protection" is notrequired.

Stability Conditions

The active power of a three-phase synchronous machine under steady-state conditions is given per phase by the equation

P = VEXd

sin � (Fig. 4.6-5)

V - stator phase voltageE - phase E.M.F. induced into the stator winding by the field current If� - load angle between phasors V and E

The equation for the power P shows that the load angle � rises if at theconstant value of P the phasor V or E sinks. However the active powerP is zero if:- the phasor V is zero (short-circuit)- the phasor E is zero (loss-of-excitation)- the load angle � is zero (no-load condition)

Acc. to the above equation the maximum value of P is reached at� = 90°. If this value is exceeded, the active power sinks and the machineis getting unstable. If the load angle is still higher than 90°, thesynchronism is lost, the speed is higher and asynchronous. The generatorthen operates as an asynchronous generator running with negative slip.The rotating magnetic field of the stator induces voltages into conductingcircuits of the rotor. Their frequency is low and is given by the slip. Theinduced voltages produce additional field winding currents and rotor eddycurrents. The output power of the prime-mover is transferred to the powersystem by the asynchronous power of the machine. The speed of aturbogenerator is higher and constant if the excitation is zero, i.e. the loss-of-excitation mode. If the field current is decreased only, than the rotormagnetic field causes speed swings. Similar swings are caused by salientpoles of salient-pole generators at zero excitation.


73

Performance of Synchronous Machines

Generally a synchronous machine operating to a bus of a large powersystem can- supply or receive an active power as generator or motor- supply an inductive or capacitive power as an overexcited or an underexcited machine.

The active power can be changed by the power of the prime-mover andthe reactive power by the field current (Fig. 4.6-1, 4.6-2). After thesynchronous machine has been synchronized and connected to the powersystem the stator current is still zero at a field current if = 1.

a) Loading by reactive power

Increasing the field current the stator inductive current rises togetherwith the reactive power. Decreasing the field current the statorcapacitive current rises. At the field current if = 0 the stator capacitivecurrent for a turbogenerator, directly connected to a power systemaccording to the phasor diagram on Fig. 4.6-1, is:

Xd I = V or I = VXd

In the p.u. system i = 1xd

At the usual value xd = 2 the current is i = 12

= 0.5

and the maximum capacitive power at the constant voltage q = 0.5.

b) Loading by active power

Increasing the power of the prime mover the active power rises (ge-nerator mode). Decreasing the power of the prime mover the activepower sinks (motor mode). At an increased active power the load anglerises; the stator current is getting capacitive if the field current is notincreased too. Under stable conditions the load angle can rise up to avalue of 90° (Fig. 4.6-4,5.6). At a load angle � > 90° the active powerof the generator starts to sink and the operation is getting unstable andthe synchronism is lost.The limit of 90° can be reached only at acapacitive current (Fig. 4.6-2).

With respect to the danger of damage caused by a loss of synchronismthe "loss-of-excitation protection" is applied to trip the generator. Thisprotection operates when the load angle value � = 90° is exceeded. Thelocus of all points with � = 90° is a circle with a diameter given by the


74

synchronous reactance Xd (Fig. 4.6-7). This circle is the limit of thesteady-state stability for generators directly connected to a power system.The load angle � = 90° can be reached acc.to Fig. 4.6-6 at variousvalues of V and E, or at V = 0 as well as at E = 0. The case with V =0 means a short-circuit on generator terminals; the case with E = 0means " loss-of-excitation". Both these points are on the axis X of thediagram on Fig. 4.6-7.

The stability of a generator-transformer unit depends on the load anglebetween the phasors of the power system voltage VS and E. The locusof the steady-state stability is then a circle given by the reactances Xdand XT of the transformer (Fig. 4.6-9).

The protection device uses a modified stability curve for generators aswell as for generator transformer units. Its operating curve (Fig. 4.6-10)

has the upper point shifted down by 12

Xd' to avoid any maloperation

during- short-circuits- out-of-step swings after a longer fault clearing.

The protection operates with a delay of 2 s to avoid any maloperation attransient conditions such as a sudden voltage change, stable swings, etc.

An integrator is used to enable a trip at stable swings at which the pro-tection device picks up, drops out, picks up again, etc.(Fig. 4.6-11).

For setting values for turbogenerators the reactance values Xd and Xd'must be known. For salient-pole generators the point with the maximumactive power is not identical with the point at � = 90°; nevertheless thedevice is often set for the detection of 90° and the reactances Xq andXd' are used for settings acc. to Fig. 4.6-3.

A typical application of the device is the detection of load angles of about90° (Fig. 4.6-13). At the reactance value Xd' = 0 the operating charac-teristic would be:- a circle touching the origin of the R-X diagram- a straight line in the power diagram.

At a real value of Xd' or XT both characteristics are circles in the R-Xdiagram.

The REG216/316 system enables the shifting of the axis of the circles andthus to set another load angle limit, e.g. 70° (Fig. 4.6-14).

The salient poles of generators produce a component of the active powerwhich is independent of the field current. Therefore the stability limit char-acteristic deviates from the characteristic of turbogenerators. The


75

diagrams in Fig. 4.6-15,16 show voltage, current and power at variousload angles and should help to better understand the stability limitcharacteristics in Fig. 4.6-17,18. For transient values at loss-of-excitationrefer to Fig. 4.6-19,20,21.

4.7. Pole Slipping Protection

Pole slipping protection has been developed to protect the generator rotorfrom disturbances causing dynamic instability. Such abnormal operatingconditions are intiated by sudden load changes which cause a loss ofsynchronisms (unstable swings).

Loss of synchronism is an onerous condition for the rotor, since it resultsin the following:- Overheating of the rotor surface by induced eddy currents- Increased mechanical stress- Possible overvoltages across the excitation winding.

The overvoltages mentioned above occur at synchronous machinesequipped with semiconductor excitation systems.

An active power of a synchronous machine is given by the formula

P = VI cos � �VEXd

= sin � (per phase)

In accordance with that the load angle � must rise if the voltage drops at aconstant power P. Whilst fast changes of E.M.S. "E" are restricted by thetime constant of eddy currents circulating in the rotor body, the voltagecan be changed suddenly by a short-circuit. Under steady-state stabilityconditions the load angle can rise only up to 90°.

When the value 90° is exceeded, the synchronism is lost and the machinebegins to slip.

A well known example of such conditions is the supply of a power systemby a generator through two lines connected in parallel. During asymmetrical short-circuit of one of the lines the voltage drops to zero andthe transfer of the active as well of the reactive power is interrupted.Because the power of the prime mover cannot be transferred to the powersystem anymore, the turbine-generator accelerates. The load angle risesup to the instant when the fault is cleared by the line protection. At a fastclearing the load angle rise is not so high and the stability is still not yetlost. After several stable and damped swings the load angle is constantagain.

The load angle rise depends on the actual turbine power, on the timeconstant �A and on the operating time of the line C.B.


76

The load angle can be calculated acc. to the following formula:

� ��

�

�� N

N

A N

PP

t1802

2 [° ; 1/s ; s ; MW ; MVA ; s]

Example:

�N = 314 1/s ; �A = 10 s ; PPN

= 0.8 ; �N = 36°

t - fault duration �N - rated load angle

t s 0.05 0.1 0.15 0.2 0.25

�� 38 43 52 65 81

After "out-of-step" the stator reactance is reduced to the value of the tran-sient reactance X'd due to the currents induced into the rotor. The currentthen varies between a minimum and a maximum value depending on thevoltage varying between the difference and the sum of voltage phasors ofthe protected generator and of the power system. The impedance givenby the generator voltage and current varies between a minimum and amaximum value too (Fig. 4.7-3, 4.7-4). If the load angle of a generatorrises the service point given by the impedance phasor moves from theright side (generator mode) to the left side (motor mode) crossing the X-axis close to the origin. This crossing is considered as a slip. The numberof permitted slips must be specified by the generator manufacturer.

After a longer fault clearing time when the load angle is greater than 90° itmay happen that the load angle rises more and more when the recoveryvoltage again enables a power transfer through the healthy line.

If the load angle � > 90° the loss of the stability is to be expected; if therating of the generator is equal or higher than 500 MW or if it is especiallyrequired by the generator manufacturer, a "pole slipping protection" is tobe applied.

Pole Slipping Conditions

Under "pole slipping" conditions the generator active power supplied tothe system varies between positive and negative values causing heavymechanical surges.

Under these conditions the rotor of the protected generator is no longersynchronous with the power system (nG � nS). The simplified behaviourof the generator and of the system can be represented by theirelectromotive forces EG, ES and the reactances. The generatoroperates with the synchronous, transient and subtransient reactanceaccording to the duration of the current-swings. With respect to the usual


77

duration of unstable rotor swings in the range up to 1s, the generator canbe represented under these conditions by the transient reactance.

The generator current can be calculated from the instantaneous voltagedifference (R.M.S.) between EG, ES and from the total reactance Xt, whichusually consists of the generator transient reactance Xd', of the generatortransformer restance XT and of the transient system reactance XS.

The phasors EG and ES rotate with slightly different angular velocities.They can also be considered such that the phasor EG does not moveand that the other phasor ES rotates with a slip angular velocity (Fig. 4.7-2).The voltage difference between EG and ES varies between a minimumand a maximum value, as does the current. The impedance measured atthe generator terminals varies between a corresponding maximum andminimum value:

Imin = E EX

G S

t

� Imax = E EX

G S

t

�

Zmax = VImin

Zmin = VImax

The measured impedance locus is a circle if EG and ES (R.M.S.) areconstant. The location of the circle is given by the location of the VT'sused for the impedance measurement. The origin of the X-R diagram isthe point where the voltage is zero and the point of the VT's location. Forthe following considerations, the usual location of the VT's at thegenerator terminals is foreseen.

Let us assume that the electromotive forces of the generator EG and ofthe system ES are of the same amplitude. The circulating current (R.M.S.)then varies between zero and a certain maximum value given by thedouble E.M.F. and by the total transient reactance. At the current zero,the measured impedance is infinite and the load angle is zero. Theimpedance point on the R-X diagram moves to the X-axis for a rising loadangle (Fig. 4.7-3). On crossing the X-axis, the load angle reaches 180°.The next point when the measured impedance is infinite occurs at a loadangle of 360°.

If the E.M.F.'s EG and ES are not equal (EG � ES), the impedancelocus is a circle. The current varies between a minimum and a maximumvalue as well as the impedance given by the terminal voltage and thecurrent.

Generally, three extreme cases are possible:

a) EE

G

S

= 1


78

circle diameter = � , the impedance locus is a straight line

b) EG = 0

full loss of excitation, the impedance locus is a point; its location isgiven by Xd'. This point is inside the circle of the "loss-of-excitation"device.

c) ES = 0

full loss of system excitation, the impedance locus is a point; its location isgiven by XT + XS .

On the R-X diagram (Fig. 4.7-3) the generator mode is drawn on the rightside of the X-axis and the motor mode on the left side. The upperquadrants are used for the overexcited mode with the ratio

EE

G

S

> 1

and the lower quadrants for the underexcited mode with the ratio

EE

G

S

< 1

E.g. at the ratio

EE

G

S

= 1

with the rising active power of a generator the measured impedance pointmoves to the X-axis and the load angle � also rises. Crossing the X-axis,the angle � = 180°.

Considering, for example the circle where

EE

G

S

� 1

the upper point on the X-axis corresponds with the highest voltage dif-ference of EG and ES out of phase and to the highest current at � =180°. The lowest point corresponds with the smallest voltage difference ofEG and ES in phase and to the smallest current at � = 0.

If, for unstable swings of a generator the load angle rises, then the X-axis


79

is crossed by the impedance point path anti-clockwise in theneighborhood of the origin. In this case, nG > nS.

If, for an out-of-step motor, the impedance point crosses the X-axis fromleft to right near to the origin, the nM < nS and the locus path is clockwise.

From the maximum and minimum values of the measured impedance thediameter D of the impedance locus circle can be calculated:

DX X XEE

EE

d T S

S

G

G

S

��

��

��

�

��

��

�

��

'

1 1

and for

EG = 0 : D = 0ES = 0 : D = 0EG = ES : D = �

According to Fig. 4.7-3,5,6 for unstable swings caused by loss ofsynchronism, the end-point of the measured impedance phasor crossesthe straight line given by Xd', XT and XS at each slip.

As already mentioned, the impedance end phasor points on the straightline Xd' + XT + XS are points where � = 180°. If a circuit breaker interruptsthe current at � = 180° then it must withstand the highest possiblerecovery voltage. Therefore it is preferable to trip e.g. at load angle 270°.This means a setting value of 90°.

The reactance characteristic is used to define two protection zones of thegenerator and of the system. The usual setting of the reactancecharacteristic enables the protection to distinguish between the slips of anoverexcited (EG > ES) and of an underexcited (EG < ES) synchronousmachine and to permit different durations of the out-of-step operation inthese two cases. For this purpose, the pole slipping device has 2 stagesNo. I and No. II with individual counters for the number of slips.

Thase stages initiate an alarm or trip as follows:

stage No. I slips at EG < ES

stage No II slips at EG > ESorslips at EG > ESandslips at EG < ES.


80

4.8. Reverse Power Protection

The reverse power device is the protection of theprime mover against motoring- generator and prime mover against overspeed.

This device is to be used for the following kinds of prime movers:- Steam turbines- Gas turbines- Hydraulic turbines, type Francis and Kaplan- Diesel motors.

Motoring is dangerous under various conditions such as overtemperatureon steam turbines, overtemperature and possibility of fuel explosion ongas turbines, mechanical stress on Diesel motors etc.

The reverse power occurs at- closed input valves of the prime mover- faulted prime mover or its supply- failured control system- shut down without tripping the C.B. or an inadvertent trip by the control system or by hand- changed power system frequency.

Aside from the protection at various motoring conditions the reversepower device is also suitable to restrict the danger of overspeed.

The motoring power of a generator driving a prime mover with closedinput valves is given by the total losses of the generator-prime mover unit.

4.8.1. Motoring

The power setting value of the reverse power device is to be selectedsuch that the device picks up at closed input valves of the prime mover.

That pick-up value is very low for steam turbines, e.g. 1 %. In order toguarantee a trip, the power setting value is usually selected according toone half of the pick-up value at closed input valves (Fig. 4.8-1).

The motoring conditions cause a danger coming not suddenly but risingslowly with time. E.g. the temperature of a steam turbine running at no-load rises according to a relatively long time constant. Therefore thereverse power device operates with a delay of e.g. 20 s for steamturbines. This delay is necessary to avoid any maloperation at suddenload swings, especially after just synchronizing a generator or if it operateswith a small load.

During repeated power swings the applied delay disables the tripping of


81

the circuit breaker, because the protection device picks-up and drops-outagain before a trip can be released by the timer. In this case an additionalintegrating timer is to be used (Fig.4.8-2). Due to its longer reset time thetrip is guaranteed also during swings. The integrator is used especially forsteam turbines, because their motoring power is small.

For ABB gas turbines two different power setting values are specified, e.g.1.5 % and 8 %.

4.8.2. Overspeed

After disconnection of a loaded generator, it is accelerated by the prime-mover; the speed rises .Subsequently the speed is decreased to thenormal value by the speed regulator. For the case of a faulted speedregulator a back-up protection is required.

In order to restrict the danger of overspeed:- the circuit breaker is immediately tripped only during dangerous faults such as short-circuits and some ground faults- the circuit breaker can be tripped during less dangerous faults after the prime mover power has been reduced to zero with the help of the control system and the trip command has been initiated by a reverse or low for- ward power device- before any shut down the C.B. is to be tripped by a reverse power device after the prime mover power has been reduced to zero.

The reverse power protection applied against overspeed should be fast.Only a short delay of 0.5 or 1 s is used to avoid any maloperation. Withrespect to a required high sensitive setting of the protection device forsteam turbines, causing an increased danger of maloperation, the trippingis initiated by an auxiliary contact, after all inlet valves of the steam turbinehave been closed. This interlocking has the task to avoid any inadvertenttripping of the loaded generator.

4.8.3. Conditions on Voltage and Current Transformers

Most steam turbines have such small losses that the power devicesensitivity is required to be very high. The measurement of such very lowpower values may be affected by phase-angle displacements of thesupplying voltage and current transformers, specially at lower powerfactors. It is therefore recommended to connect the reverse power deviceonly to protective or measuring transformers of a sufficient accuracy andof a very small phase-angle displacement.

4.9. Inadvertent Energization Protection

The inadvertent energization protection is applied to protect the primemover-generator unit against:


82

- overcurrent and overtemperature- winding dynamic stress- mechanical stress of rotor and bearings.

This protection is also known as standstill or dead machine protection.Caused by a failure of the control system an inadvertent energization canhappen at:- standstill- start-up- running without or with voltage.

A sudden energization at standstill is most dangerous. It causes a high cir-culating current and the highest dynamic as well as mechanical stress.The danger of damage is the highest for bearings at low oil pressure.Therefore a fast trip is required.

The circulating current is given at a standstill energization by the generatorreactance Xd'', the transformer reactance XT, the power systemreactance XS and by an active current component supplying generatorand prime-mover mechanical losses. Monitoring of the stator current or ofthe power can be applied for the protection. Naturally the appliedprotection device must be stable during normal operation as well as duringexternal faults and transients.

With respect to the applied delays of the generator overcurrent protectionand of the reverse power protection, these devices cannot be consideredas suitable for an inadvertent energization protection, and anotherprotection device is to be applied. There are two different protectiondevices applied, operating either with overcurrent or power.

The overcurrent method uses an instantaneous overcurrent devicereleased by an undervoltage device. A maloperation of the overcurrentmethod at nearby short-circuits is prohibited by a pick-up delay of theundervoltage device.

The power method uses an instantaneous power device detecting apower received to supply the desired reactive power and the active powergiven by total losses. A power setting should be higher than the setting ofthe reverse power device, e.g. for 10 % at steam and 15 % for gasturbines. A suitable phase angle setting is around 60°.

4.10. Voltage Unbalance Protection

The voltage unbalance protection is used to detect any loss of an ACvoltage desired for the supply of some protection devices which couldmaloperate due to a loss of an AC voltage; it is the minimum impedancedevice, the minimum reactance device, the reverse power device and theundervoltage device if applied.


83

The voltage unbalance device operates by monitoring the difference bet-ween the voltages of two sets of VT's. A detected voltage difference canbe caused by:- a blown fuse or open miniature C.B. on the L.V. side- VT faults- a blown fuse on the M.V. side.

Features of the voltage difference monitoring:- Restricted influence of the M.V. level fluctuation (at continuous as well as short-time conditions)- Phase-selective detection.

If this protection device operates, it has to block the above mentionedprotection devices in order to avoid maloperation.


84

5. Protection Current Transformers

Current transformers applied for the protection must be able to transfer theprimary current to the secondary side also at high fault currents with asufficient accuracy.

The current transformers, which must not saturate under given faultconditions, are to be designed such that a certain maximum flux density,allowed with respect to the maximum permitted current error, usually being1.8 Vs/m2, will be not exceeded. The CT's are to be designed according tothe overcurrent factor n, given by the ratio of the maximum magnetic flux tothe flux at the rated current. This factor n is calculated by the ac faultcurrent component i (p.u.) and by the factor k corresponding to the dc faultcurrent component:

n = k i

The factor n depends on

- the frequency - the AC fault current component i (p.u.) - the time constants of the faulted ciircuit � and of the CT secondary side circuit �2 - the time t.

The current transformer is to be specified according to the highest faultcurrent. On the generator side this is the 3phase short-circuit current and onthe main transformer HV side the line-to-ground fault current (refer to Table2-VI). The fault current comprises an ac and a dc component. The accomponent is to be calculated by transformer reactances, generator sub-transient and inverse reactances. At the calculation of fault currents usedfor the specification of CT's the damping of the dc component is consideredbut no damping of the ac component. For examples of the ac component atH.V. faults on a generator-transformer unit refer to Table 2-VI. The dccomponent depends on the instant of the fault. It is approximately zerowhen the fault occures at the peak of the voltage sinus wave and it is thehighest for faults at a voltage equal to zero. The damping of the dccomponent is involved in the calculation by means of the time constant forthe dc component �.

The fault current during a short-circuit on a certain unit supplied by an acvoltage can be calculated from the equation

V 2 sin ( t + ) = R i + L didtF

F� �

where


85

V phase voltage of the supplying power system (RMS)iF instantaneous fault current� angular velocityt time� angle of the voltage curve at the time t = 0.R, L resistance and inductance of the faulted unit (transformer,generator)

The fault current given by the solution of the above equation is:

i = K e + VZ

sin ( t +F

-t-� � � ��

2

where Z is the impedance given by R and L and � is the time constant of thefaulted unit (generator, transformer or both). The factor K is to bedetermined for the current iF at the time zero. If the current iF is zero at theinstant of the fault we get

i = K + VZ

sin ( = 0

K = -VZ

sin (

F(t=0) -2

2

� �

��

�

�

and the fault current is

i = VZ

sin ( t + - e sin ( F ( )- -t2

� � � ��

The dc component depends on the angle (�-�). This component is zero atthe angle � = �. The fault current is then:

i = VZ

sin tF2

�

or using the RMS value of the fault current IF

I = VZF

i = I 2 sin tF F �

A fault current limited especially by a resistor, as is the case of a groundfault on a resistance grounded transformer, is given by the same equation.

The full dc component is developed at the angle (�-�) = -90°. It may be e.g.at � = 0 and � = 90°. In this case the fault current is:


86

i = V 2Z

sin ( t - 90 ) - e sin (-90 ) F

-( )

t�

��

i = V 2Z

e-

- cos t F ( )t� �

or using the RMS value of the ac component of the fault current

i = I 2 e - cos t F F ( )-t�

�

5.1. Current Transformers with Closed Magnetic Core

Fault Current without a DC Component

The desired overcurrent factor n is to be calculated from the function of themagnetizing current, given by the difference of the CT primary andsecondary ampere turns. The applied angles are � = �. For the calculationthe following basic equations can be used:

i ii = I 2 sin ( t)

0 = (R + R ) i + L didt

M didt

1 = F

1 F

2 B 2 22 1

�

�

where

i1, i2 CT primary and secondary instantaneous currents

R2 CT secondary winding resistance

RB burden resistance (the burden by static protection systems is resistive)

L2 CT total secondary winding inductance

M CT mutual inductance

At first we have to calculate the CT secondary current; applying the timeconstant of the CT secondary circuit:

�2

= LR R

2

2 B�


87

didt

I 2 cos t

0 = didt

+ 1 i - ML

I 2 cos t

i

I 2 = e K + M

L e

1+ ( ) ( 1 cos t + sin t)

1F

2

2

22

F

2

F

2 2

22

2

-t

2

t2( )

� � �

� �

�

�

� � �

�

��

� �

� 2

At the instant t = 0 the secondary current is also zero. The constant K is

K = ML

-

1+ (

and the current i

i

I 2= M

L 1+ (

e - ( cos t + sin t)

2

-t

2

2

2

2

2

F 2

2

22

2( )

�

�

�

�

� � �

�

�

�

�

�

�

�

�

2

�

With the help of the equation for CT ampere turns

N1i1 = N1i1exc + N2i2

where N1 and N2 are numbers of primary and secondary winding turns, weget the equation for the instantaneous magnetizing (excitation) currenti1exc:

i

I 2 = 1

I 2 (i - N

Ni )1exc

12

12

F F

Neglecting the secondary winding leakage inductance L2L, we getL2 = L2exc and therefore the ratio of the secondary self-inductance L2 tothe mutual inductance M is

LM

= NN

2 2

1

The value of L2exc can be calculated from the CT magnetizing current. Ifthis current is still unknown, its value can be estimated according to thecurrent error allowed by the used CT class. E.g. at the class 5P20 theallowed magnetizing current is up to 5% at 20 I1N.

For the instantaneous magnetizing current i1exc we get :


88

i

I 2 = 1

1+ ( sin t + (e - cos t )1exc

F2 2

( )-t

2

�

� � �

�

�

�

�

2

Applying the factor k corresponding to the influence of the dc component ofthe magnetizing current

k = e - cos t

-t

2�

�

we get a simplified formula for the instantaneous value of the magnetizingcurrent at a fault if ��2 >> 1:

i

I 2 1 e - cos t) = 1 k 1exc

F 2 2

-t

��

�

��

��

2

The magnetizing current at the rated load is

I = EX

= E N

N

X NN

= (R R I N

NL

1 I1excN1N

1exc

2N

2exc

2N

2exc1N

1

2

1

2

2

2 B2

1

2

� �

�

� � �

��

�

�

when we neglect the magnetizing current in the equation for CT ampere-turns for the calculation of the ratio of the primary and secondary currentsand put N1 I1N � N2 I2N. The currents I1N and I2N are rated currents.

For the p.u. value of the magnetizing current at rated load (refer to Fig.3.2-7)

i = II

1excN

1excN

1N 2

�

� �

is also the CT current error of the CT primary rated current I1N under theassumption that the CT turn-ratio is correct. According to this formula thetime constant of the secondary circuit �2 must be equal or higher than 318ms at magnetizing currents up to 1% and at 50 Hz. It means that in thiscase the product ��2 >> 100. According to the class 5P20 the allowedmagnetizing current is 1% at the rated current and at the rated load.

The p.u. value of the ac component of the fault current IF:

i = II

F

1N


89

The instantaneous magnetizing current at a fault and based on theamplitude of the magnetizing current at the rated load i1excN:

i

I 2 = I e - cos t

I = k i1exc

1excN 1N

F

2

-t

2( ) 2

�

�

�

�

��

The ratio of the instantaneous magnetizing current at a fault to the peakvalue of the magnetizing current at the rated load causes an increasedmagnetic intensity, important for the design of the CT magnetic core. Thisratio is called overcurrent factor n, used for the specification of protectionCT's. This ratio:

n = i

I = k i1exc

1excN 2

According to the equation with factor k as a function of the time, the factor kreaches values up to 2 in the worst case with a large time constant�2 . Therefore the CT's are to be specified at least with the factor k = 2 ,also in cases with no dc component of the fault current. This is e.g. the caseof ground fault currents on a resistance grounded power system.

Fault Current with a DC Component

The CT primary current i1 comprising a full dc component is:

i = I 2 (e - cos t)1

-t

F�

�

Similarly as at fault currents without any dc components we get now:

iI 2

= -ML -

e - e + ML

)

1 + ( ) e +

+ ML

( )

1 + ( ) 1 sin t - cos t

2

F 2

-t

2

2

22

22

2

2

22

22

2

( ) )

( )

2

-t -ttt t

tt

tt t

t t t(ww

ww w w w

2

and for ��2 >> 1:

i

I 2 -M

L

- e -

- e + cos t - 1 sin t 2

F 2 2

2

2

( )

2

-t-t

�

�

� �

�

� � �

�

�

�

�

�2

The instantaneous magnetizing (excitation) current given by the differenceof primary and secondary ampere turns:


90

i

I 2 =

-e -

- e -

( )

1+ ( )e -

- (

1+ ( ) 1 sin t - cos t - cos t

1exc

2

-t-t -t

F

2

2

2 22

22

2

22

22

2

( )

�

� �

�

� �

�

�

�

� �

�

� ��

�

�

� �

� � �

�

The simplified formula for this current at (��2)2 >>1:

i

I 2

- e - e - 1 sin t 1exc

F 2

-t-t

22

(�

�

� � �

�

�

��

�

The approximate peak value of the magnetizing current at each negativepeak value of the sinus wave and at (�t)2 >> 1:

i

I 2 1

- (e - e ) + 11exc

-t-t

F 2

2

2

2( )�

�

�

�

��

� �

�

�

The p.u.value of the magnetizing current can be calculated according to theFig.3.2-5

I EX

1 I 1excN1n

1exc =

2

1N�

��

i

I 2 I

I

- (e - e ) + 1 1exc

F

1excN

1N

2

2

-t-t

2( )�

��

� �

�

�

i

I 2 (e - e ) + 1 i where is i = I

I1exc

1excN

-t-t

( )2

2

2 F

1N�

�

��

� �

�

�

With the factor k the overcurrent factor n is:

k = -

e - e + 1 n = i

I 2 = k i2

-t-t

1exc

1excN 2

2��

� �

� ��

�

For the factor k of CT's with a large time constant �2 >> � we get asimplified formula

k (1 - e ) + 1

and for the time t =

k + 1

-t�

�

�

�

�

�

�

�


91

For the comparison of the curves for the factor k according to above threeformulas see Fig. 5.1-1.

For curves of the desired factor k as a function of time t refer to Fig. 5.1-2.These curves have a peak value

k = + 1m

2

( ) 2� ��

�

�

� ��

at the time

t = ln m2

2 2

-

( )� �

� �

�

�

For the curves of the factor km as a function of the time constant �2 at giventime constants � see Fig.5.2-3.

Often very large and expensive CT's are specified as a result of aboveequations. Therefore often a certain saturation is allowed or the timewithout saturation reduced in order to reduce the factor n (refer toFig. 5.1-1).

According to above equations the time constant �2 depends on the CT no-load current (magnetizing current). For the class 5P20 the allowed limits are1 and 5% for the current error at I1N and 20 I1N. Nevertheless the actualvalues are usually smaller. Especielly at CT's for the generator current theactual no-load currents and corresponding error values are very low, e.g.0.002 due to the high number of ampere turns. The current error of a CT for10 000 A may therefore be on the level of 0.2%. The value of �2 is thenmuch higher e.g. 1.6s at 50 Hz or more. In such a case it is possible tocalculate the factor n according to approximate formulas as considered inChapter 3.2.4.2.

Estimation of CT Data for Steady-State and Transient Conditions

For each project it is necessary temporarily to specify the protection CT'swith respect to steady-state and transient conditions. Thereforeapproximate values of the magnetic core cross section and of theresistance R2 are desired. The value of the core cross section determineswhether the CT can be realized and built in; at CT's for generators thisvalue should not exceed 100 cm2. The value of the resistance R2 isrequired in order to check the transient performance of the CT's at high faultcurrents.

For the development of the desired formulas for the factor n the followingequations can be used. The electromotive force E2 desired for the cal-culation of the CT


92

E2 = 4.44 f N2 B A [V; Hz; Vs/m2;m2]

or

E2 = 4.44 f N2 B A 10-4 [V; Hz; Vs/m2;cm2]

The cross section A of the magnetic core with the sides a

A = a = wire length4 1,25 N

= R s

5 N cm m ; m / mm mm2

x x2( ) ( )

2

2

2

2 2 2 2 S

��

The factor 1.25 is a safety margin, because the actual length of one turn islonger than the size of the magnetic core. The value s is the wire crosssection.

Applying the specific copper conductance � = 50 S m/mm2 we get:

A = 10 R sN

6 2

2

2( )

and using the typical values of the wire cross section we get

for I2N = 1 A and s = 0.6 mm2: A = 0.36x10 RN

6 2

2

2( )

I2N = 5 A and s = 3 mm2: A = 9x10 RN

6 2

2

2( )The protection CT's have to transfer the primary current to the secondaryside with a certain accuracy up to the highest current specified by theovercurrent factor. With respect to the required accuracy the magnetic fluxdensity should not exceed the value 1.8 Vs/m2. Therefore we takeB = 1.8 Vs/m2

and we get

E = 4.44 f N 1.8 10 R sN

10

E 800 f (R s)N

= n (R I + V )

2 26 2

2

2 -4

22

2

22 2N 2N

x ( )

�

The equation for the resistance R2

(R f sN

- R (n I - n V22

2

22 2N 2N� � = 0

The solution of this equation gives


93

R =

n I2

+ n I2

+ 800 f s n VN

800 fN

s2

1N 1N 2 2N

2

2

) 2

2

�

1. Equations for f = 50 Hz

1.1 Secondary rated current I2N = 1 A and s = 0.6 mm2

n = optional value :

R =

n2

+ n2

+ 14400N

n V

14400N

2

2

)2

22N�

A = 0.36 10 RN

x 6 2

2

2( )

n = 20 :

R = 10 + 100 + 288

N V

360N

102

2N

2

32

103

A = as above

1.2 Secondary rated current I2N = 5 A and s = 3 mm2


R = 2.5 n + 6.25 n + 360

N 10 n V

360N

102

2

3

2

2

32N

A = 9 10 RN

x 6 2

2

2( )

n = 20 :

R = 50 + 2500 + 7200

N V

360N

102

2

3

22N

A = as above

2. Equations for f = 60 Hz

2.1 Secondary rated current I2N = 1 A and s = 0.6 mm2


94


R =

n2

+ n2

+ 17.28N

10 n V

17280N

2

2

( ) 2

2

32N

A = 0.36 10 RN

x 6 2

2

2( )

n = 20 :

R = 10 + 100 + 345.6

N V

17280N

2

2

22N103

A = as above

2.2 Secondary rated current I2N = 5 A and s = 3 mm2


R = 2.5 n + 6.25 n + 432

N 10 n V

432N

102

2

3

2

2

32N

A = 9 10R

Nx 6 2

2

2� �

n = 20 :

R = 50 + 2500 + 8640

N 10 V

432N

102

2

3

2

32N

A = as above

With the help of the above formulas the secondary winding resistance,losses and the magnetic core cross section were calculated as example forthe following CT's with four various current ratioes.

CT ......../5 A 50 Hz 5P20 30 VAwindow, bar or bushing-type with only one primary turn (N1 = 1).

item No.

CT ratio [A]

R2 [�]

R2 20I2N [V]

E2 [V]

�PCu2 [W]

A [cm2]

1 100/5 0.084475 8.4475 128.45 2.11 160 2 1000/5 0.28747 28.747 148.75 7.18 18.6 3 10000/5 1.1402 114.02 234.02 28.50 2.9 4 50000/5 3.6829 368.29 488.29 92.07 1.2

Remarks to

item No. 1: - Small losses


95

- Magnetic core cross section too great;possible help: - reduction of the rated output (especially

important for cable summation CT's providedfor a sensitive ground fault protection)

- prefer a wound-type CT, which hasa higher number of ampere turns

item No. 2 : - Losses and size of the magnetic core are suitable- If a much higher rated output is required, - prefer a H.V. CT with 2 or 4 primary winding

turnsitem No. 3: Acceptable losses and a small size of the magnetic core.

Using an increased magnetic core a much higher rated output can be reached. For details refer to the next table.

item No.4 : High losses and a very small size of the magnetic core. Such very high losses must be excepted. Try to restrict the number of proposed CT's and apply them for various protection,measurement and control circuits.

CT's with the ratio 1000/5 A, number of primary turns N1 = 1, number ofampere-turns N1 I1N = 1000 A at various rated outputs:

SN [VA]

V2N [V]

20 V2N [V]

R2 [�]

R220I2N [V]

E2 [V]

�PCu2 [W]

A [cm2]

15 3 60 0.2124 21.24 81.24 5.3 10.15 30 6 120 0.2875 28.75 148.75 7.1 18.60 60 12 240 0.3940 39.40 279.40 9.8 34.90 100 20 400 0.5000 50.00 450.00 12.5 56.25

According to the above table a large magnetic core would be necessary forratings 60 and 100 VA. Using a higher number of ampere turns the corecross section would be smaller. For the influence of several primary windingturns refer to the next table.

CT's with the ratio 1000/5 A , at various numbers of primary turns N1 . Therated output is 100 VA, the rated output voltage V2N = 20 V, 20 V2N = 400V.

N1 N1 I1N N2 R2 [�]

R220I2N [V]

E2 [V]

�PCu2 [W]

A [cm2]

1 1000 200 0.5000 50.00 450.00 12.5 56.25 2 2000 400 0.7245 72.45 472.45 18.1 29.5 4 4000 800 1.0604 106.04 506.04 26.5 15.8

It is obvious that the rising number of ampere-turns N1 I1N enables todecrease the cross section of the magnetic core "A". Unfortunately onlysome types of H.V. current transformers have the possibility of severalprimary turns.


96

Sometimes it is necessary to check the CT phase displacement. Using theFig. 3.2-5 it is possible for small values of the phase error to write

��

arc tg 1 rad; 1 / s; s2

�

and

= 3602

60 1 1000 = 10 943 min; 1 / s; ms 2 2

��

E.g. application of a time constant �2 = 60 ms means a phase displacementof 182 min.

Note

The calculation of the resistance R2 and of the magnetic core cross sectionA described above is to be applied for preliminary values only, desired atthe begin of the project. The final values are to be asked from the CTmanufacturer.

5.2. Linear Current Transformers

Sometimes an application of linear current transformers with an air-gap isrequired in order to improve the transfer capability of the ac currentcomponent under transient conditions and thus to reduce the CT size.

The transfer of the less important dc component to the secondary side isreduced.

The features:

Advantage: - much smaller cross section of the magnetic core- smaller copper losses

Disadvantage: - higher magnetizing current.- secondary winding leakage inductance not negligible

Requirement: The same time constant of the secondary circuits of §both CT's for the differential protection if possible.

Remark: The application of linear CT's is not favourable for a small number of ampere turns.

For linear CT's the same basic equations as those developed for CT's withclosed magnetic core may be used, only the leakage inductance of thesecondary winding is not negligable.

At given overcurrent conditions the desired overcurrent factor n for linearcurrent transformers is much smaller than this factor for currenttransformers with a closed magnetic core. The linear CT's operate with areduced magnetizing inductance. Due to the reduced magnetizing


97

inductance the core cross section sinks. Naturally the reduced magnetizinginductance causes an increased magnetizing current. Depending on thereduction of the magnetizing inductance the cross section of a linearmagnetic core may be e.g. 3 up to 5 times smaller than the cross section ofa closed magnetic core.

At CT's for differential protection the increased magnetizing current must bekept so small, that a possible difference between magnetizing currents ofboth CT's supplying this protection does not cause a maloperationespecially at external faults. Therefore a maximum magnetizing current of3% at CT's for differential protection should not be exceeded. For otherprotection funtions supplied by one CT only the magnetizing current is lessimportant; it is possible to accept its values of up to 5 or 6% and thus toreach a reduction of the magnetic core up e.g. up to 5-times.

Determination of Desired Data for Linear CT's

At the begin it is necessary to select the allowed current error. At protectionCT's with a closed magnetic core, the allowed current error is e.g. 1% forthe class 5P20 in the range of normal service currents; this current errormay rise up to the value of 5% at the current 20 I1N. In contrast theselected current error of a linear CT is a constant value e.g. of 5% in thetotal range of currents for steady-state as well as transient conditions,specified by the overcurrent factor n, it means in the total range of CT linearcurrent transfer capability.

Therefore the allowed current error should be selected according to theapplication of the CT. For the CT's used for overcurrent or distanceprotection current errors up to 5 or 6% may be allowed. In contrast to theprotection operating with a difference of currents such as the differentialprotection lower values of the allowed current, e.g. 3% should be preferred.

With the help of the selected current error, given by the magnetizing currentI1excN according to Fig. 3.2-5, the CT secondary circuit time constant �2can be calculated:

� � �

� � �2

1N

1excN 21 1N

1 1excN 2excN

1 II

or 1 N IN I

or 1i

� ��

E.g. at a selected current error of 0.01 (iexcN = 0.01 or 1%) and � = 314 1/sthe time constant �2 = 0.318 s or 318 ms. For several examples refer to thefollowing table.Magnetizing currents at various values of the time constant �2 (at 50Hz):

�2 [ms] 60 90 120 150 180 318.3 3 183 31 830iexcN[%] 5.3 3.54 2.6 2.12 1.77 1 0.1 0.01


98

According to the above recommendation of allowed current errors the CT'sfor a differential protection can be specified by a time constant �2 = 120 msand the CT's for other protection functions by �2 = 60 ms.

The required air-gap g is given by the ratio of magnetizing ampere-turns(peak) n N1I1excN �2 to the magnetizing force H or to the flux density B:

g = n N I 2H

= n N I 2 0.8 B

10 m; A; Vs / m

or

g = n N I 20.8 B

10 mm; A; Vs / m

1 1excN 1 1excN -6 2

1 1excN -3 2

According to the above formulas the lengt of the air-gap is proportional tothe number of ampere turns. The air-gap is quite short at low values ofAmpere-turns and it is long at high values of ampere turns.

The air-gap length at various ampere turns:

I1N [ A ] 500 1000 10 000 20 000 40 000g [ mm ] 0.5 1.0 5.0 10 20

It is evident, that short air-gaps are difficult to be adjusted. The applied air-gap or gaps cause an irregular distribution of the magnetic in the magneticcore. The result is a magnetic flux leakage and the value of the secondaryleakage inductance L2L is not longer negligible. In order to keep themagnetic flux leakage as low as possible, the length of the air gap must bemuch smaller than the size of the magnetic core cross section. Therefore itis sometimes necessary to split the calculated value of the gap into severalshort air-gaps. This is especially the case for CT's for very high currents, asthe above table shows.

Specification

The formulas developed for CT's with a closed magnetic core andnegligeable secondary winding leakage inductance cannot be used forlinear CT's. Due to the applied air-gap the secondary leakage inductance isnot negligible. The formula for the calculation of the inner E.M.F. E2 as thesum of V2N and R2 I2N is not longer correct. The calculation of thesecondary winding leakage inductance L2L is not simple and should bemade by the CT manufacturer. Therefore the cross section of the magneticcore cannot be calculated and consequently the value of the secondarywinding resistance cannot be estimated.

Without any knowledge of L2L and R2 it is better to specify the transientperformance of the required linear CT's by:


99

- the actual burden (given by connection leads and by the protection system)

- the maximum fault current (ac component in RMS value)- the time constant of the faulted circuit- the percentage of the dc component which should be considered- the desired time constant of the CT secondary circuit and its tolerance.

Naturally the secondary circuits of both CT's for a differential protectionmust be specified by the same values of the time constant.

Conclusion

Although the linear CT has important advantages, its application should berestricted with respect to disadvantages and possible maloperations of thedifferential protection caused by differencies between both CT's.


100

6. FIGURES AND TABLES

Fig. 1-1 Generator protection.Single-line diagram acc. to ANSI standardIF - interturn fault


101

Fig. 1-2 Generator protection.Single-line diagram acc. to IEC standard


102

Item Protection Device ANSIdevicenumber

1 Overcurrent 50;5151 V

2 Overload 49

3 Unbalanced load 46

4 Voltage protection 59; 6027

5 Overexcitation 24

6 Frequency 81

7 Loss-of-excitation 40

8 Pole slipping 78

9 Standstill 67

10 Reverse power 32

11 Differential 87

12 Buchholz or SPR*) 63

13 Minimum impedance 21

14 Interturn fault -

15 Stator ground fault 59 GN64 GN

16 Rotor ground fault 64 F

17 Transformer ground fault 51 NT87 NT

18 Breaker Failure 51/62 BF

*) SPR = sudden pressure relay

Table 1-I Generator and transformer protection devices


103

Fig. 1-3 Possible kinds of faults on a generator stator windinga) Short-circuitb) Interturn faultc) Ground fault


104

Rating [MVA] <5 5-20 20-50 50-200 200-500 500>

I > / U < 51V x x x x x x

� I> (overall) 87GT x x x (x) x

�IG > 87G x x x x

Z < 21 (x) x x

Interturn fault - x x

95% Statorground fault

59GN (x) x x x x

100% Statorground fault

64GN x

Rotorground fault

64F (x) x x x x

��stator > 49 (x) x x

��rotor > 49F (x) x x

I2 > def. time 46 (x) x inverse x x

U > 59 x x x x

U < 27 x

�> def. time 24 (x) x inverse x

f >f <

81 x x x

Loss-of-excit. 40 x x x

Pole slipping - (x) x

- P > 32 x x x x

Inadvertentenergization

50/27or 67

(x) x

Table 1-II Generator protection x = typical application(x) = non-typical application


105

Rating[MVA]

� 0.5 0.5 � 1 1 � 2 2 � 5 5 � 10 10 � 100 > 100

Fuses � - - - - - -

Buchholz - - � � � � �

OvercurrentI >

- � � � � � �

OvercurrentI >>

- - � - - � �

Overload�� >

- - - - - � �

Differentialcurrent �I >

- - - � � � �

Ground faultIE >

- - - - � � �

Restrictedground fault

�IE

- - - - - � �

Overexcitation� >

- - - - - - �

Overvoltage *)U >

- - - - - � �

Table 1-III Transformer protection*) only at tap-changer transformers


106

HEST 905 011 FL

12

3

4

5

6

7

8

9

10i

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

isc

iN

t

Fig. 2-1 Transformer-Kurzschlussstromuo = umax und xT = 0,1


107

Fig. 2-2 Generator-short-circuit current


108

Fig. 2-3 Field discharge circuit breaker


109

U = Uarc + Ri + = L didt

current suppression:

didt

U U RL

arc i�

� �

at the instant t = 0 is i0 = UR

and

- didt

ULarc

� or by the time constant T

- didt

URT

arc�

the instantaneous current

I UR

e U UR

arc -tT arc

� �

�

the arcing time

t T ln UU Uarc

arc

arc

�

�

Fig. 2-4 Current suppression by dc circuit breaker


110

Ib1

Ia1

Ic1

Ib2

Ia2

Ic2

bc

a

Ib2

Ib1

Ic2

Ic1

Ib

Ic

IcEa

IbUbUc

Uc1 Ub1

Ua1

Ub2 Uc2

Ua2

Ua2Ea

Ub Ub1Uc1 Uc

Ub2Uc2

Ua1

Fig. 2-5 Line-to-line faultsymmetrical components


111

Ic2

Ia2

Ib2

Ib1

Ia1

Ic1

c

a

b

Ia0Ib0Ic0

Ia0Ia2Ia1

Ia

Ic1Ic2

Ib2 Ib1

Ib0

Ic0

Ua1

Ua0

Ua2

Ub1

UbUc

Uc1Uc2

Uc0 Ub0

Ub2

Ua1Uc2 Ub2

Ua2Uc1 Ub1

Ua0Ub0

Uc0

Ia

UbUc

Ua = 0 Ib = Ic = 0

Fig. 2-6 Line-to-ground faultsymmetrical components at x1 = x2 = x0


112

Ia0Ia2Ia1

Ia

Ic1Ic2

Ib2 Ib1

Ib0

Ic0

Ib1

Ia1

Ic1

Ic2

Ia2

Ib2

c

a

b

Ia0

Ib0Ic0

Ub2

Uc

Uc1

Uc2Ub1

Ub

Ua1 Ua2

Uc2 Ub2

Ua2Uc1 Ub1

Ua1

IaIb = Ic = 0

UbUa = 0Uc

Fig. 2-7 Line-to-ground faultsymmetrical components at x1 = x2 and x0 = 0


113

2

output

2

output

1

input

1

input

T

T

Fig. 2-8 Asymmetrical short-circuit of transformerswith the connection group Yd and Dd


114

output

T2

input

T1

Fig. 2-9 Line-to-ground fault of 2 transformers connected in parallelThe neutral point of the one transformer is solidly grounded,the neutral point of the other transformer is ungrounded.


115

T

T

T

Fig. 2-10 Asymmetrical short-circuit currents of thegenerator - generator transformer unit


116

Ia = Ib = Ic

a

c b

c b

a

a

bc

a

c b

Ib = Ic

EX1

Ub = Uc

Ea

X1 + X2

��3 E

X1 + X2

X2

X1 + X2

2X2

X1 + X2 + X0

3E

X1 + X2 + X0��3 E

X1 + X2 + X0��3 E

2X2 + X0

X1(X2 + X0)+ X2X0��3 E

EX1(X2 + X0)+ X2X0

3X2

X1(X2 + X0)+ X2X0

3X2 X0 E

Ia

Eb = Ec

Ebc

Ib = Ic

I0

Ea

E

E

Table 2-I Formulas for the calculation of short-circuitcurrents and voltages


117

I1

EX1

X1 + X2 + X0

E

I2

X1(X2 + X0)+ X2X0

X2 + X0 E

X1(X2 + X0)+ X2X0

- X0 E

X1(X2 + X0)+ X2X0

- X2 E

X1(X2 + X0)+ X2X0

X2 X0 E X1(X2 + X0)+ X2X0

X2 X0 E X1(X2 + X0)+ X2X0

X2 X0 E

0

X1 + X2 + X0

EX1 + X2 + X0

E

X1 + X2

EX1 + X2

-E 0

0

I0 E1 E2

E 0

E0

0

0EX1 + X2

X2 EX1 + X2

X2

X1 + X2 + X0

X2 + X0 EX1 + X2 + X0

-X2 EX1 + X2 + X0

- X0 E

Table 2-II Formulas for symmetrical components at short-circuits


118

A

C B

C B

A

A

C B

A

BC

phase subtransient current

transient current

steady-state current

A

B

C

6.7

6.7

6.7

4.0

4.0

4.0

1.25

1.25

1.25

0.0

2.0

2.0

3.33

0.0

0.0

0.0

1.8

1.8

C

C

C

B

B

B

A

A

A

0.00.0

5.8

5.8

7.5

4.3

4.3

7.2

7.2

0.0

0.0

0.0

0.0

0.0

0.0

6.0

4.9

4.9

Table 2-III Generator short-circuit currents

x''d = 0.15 x'd = 0.25 xd = 2.0 x2 = 0.15 xo = 0.1

ifmaxifo

= 2.5


119

xT0

1.0

0.1

10

10

8.66

8.66

1xT

10 10

2.5 8.69

� 32 xT

32 xT + xT0 x2

T+2 xT xT0

� 3x2

T+ xT xT0+x2T0�

Table 2-IV Transformer short-circuit currentsat connection groups Yy, Dy and YdyxT = 0.1


120

10

10

8.66

8.66

1xT

� 32 xT

Table 2-V Transformer short-circuit currentsconnection group Yd and DdxT = 0.1


121

1 x''d + xT

generator generatortransformer transformer

subtransient current steady-state current

4.0 4.0

3.46

5.02.89

4.0

2.0 *

3 x''d + x2 + 3xT

2.5 x 3 xd + x2 + 3xT

2.5 x ��3 xd + x2 + 2xT

2.5 xd + xT

1.19 1.19

2.13 1.84

1.77 3.06

x''d + x2 + 2xT

��3

Table 2-VI H.V. side short-circuit currents of a generator-generator transformer unit

x''d = 0.15, x2 = 0.15, xd = 2.0, xT = 0.1, ifmax

ifo = 2.5

* one generator phase is loaded by i = 4.0 and the other by i = 2.0


122

IIN

2

1

0 t

1.5

1.3

2.0

Min through faultcurrent

Possible currentsetting range

Max servicecurrentIN

Fig 3.1-1 Generator overcurrent protection


123

i

0 t

0 t

Startsignal

u

0 t

0 t

Delay Trip

Fig. 3.1-2 Generator current, voltage and signals of the protection device 51 V.


124

IIN

10

1

0 t

1.5

1.3

8.66

Max through fault current ormax. inrush current

Possible setting range ofthe definte time stage

Max service current

IN

50 43

15

10

Min. HV fault current

Possible setting range ofthe instantaneous stage

Min through fault current

Fig. 3.1-3 Overcurrent protection of a generator transformer


125

0.5 ... 1 1.67 ... 10 > 10sN (MVA)

10 ... 60

0.2 ... 1.20.16 ... 0.2

8 ... 10

t (s)

t (c) 60 ... 3600

1.2 ... 72

0.5 1 5 10

10

50

9101416

sN (MVA)

I0 peakIN

Fig. 3.1-4 Transformer magnetizing inrush currentpossible inrush current values, duration between the fulland half peak value in cycles and seconds


126

0 t

IN �2

IN �2

IN �21.5

I

ts

Fig. 3.1-5 Suitable settings at a certain level of the inrush current.


127

Minimum fault current

Current settingRated current

~

Power system 1000 MVA 10 %

T 100 MVA 10 %

AT 10 MVA 10 %

50.1

51.1

50T

51T

51.2

G

866

83

2.0

1.0

51.1 51.2

7.8

2 IATN

IATN

7.8 IATN

ITN ��10 IATN

10

15 51T 50.115 IATN� 1.5 ITN

83 IATN� 8.3 ITN

866 IATN� 86.6 ITN

100 IATN� 10 ITN

150 IATN � 15 ITN

150 50T

Fig 3.1-6 Transformer overcurrent protection.Fault current level and current settings


128

T

I>> I>

t

50 T 51 T

G ~

I>> I>

t1

50 51.1

51.2

I>

t2

AT

IIATN

150 IATN

0 0.5 1.0 1.5 2.0 2.5

50 T

50.1

51.151.22 IATN

51 T

15 IATN

IATN

t (s)

Fig. 3.1-7 Grading of overcurrent device settingsfor machine transformer T and auxiliary transformer AT.


129

� I>

G

� I>

G G

Fig. 3.2-1 Principle of differential protectionat internal and external faults.


130

iN

is

0 t

1

i

0 t

1

i

0.5

1.5is

ts

iN

Fig. 3.2-2 Comparison of overcurrent- anddifferential device operating curves.


131

0 0.1 0.2 0.3 0.4 0.5

1.3

1.2

1.1

U

iexc

U

Fig. 3.2-3 Transformer excitation current as a function of voltage.


132

0,75

0,5

0,25

I�IN

IN

IH0 0,5 1 1,5

g b

Fig. 3.2-4 Operating characteristic of the transformer differential protection forhigh through-fault currents


133

I1 I2Iexc

E1=E2

input

R1 L1L

L1 exc

R2L2L = 0

U2

Phasors for L2L = 0 Z = R

R2I2

U2=R I2Iexc

E1 = E2

I1I2

�

�

Z

Fig. 3.2-5 Equivalent ct circuit at N1 = N2

Used symbols:

I1, I2, E1, E2 primary and secondary currents and E.M.F'sU2 secondary voltageIexc., L1exc. excitation current and inductancesR1, R2 primary and secondary resistancesR, Z burden resistance and impedanceL1L, L2L leakage inductancesN1, N2 number of primary and secondary winding turns� phase displacement


134

H.V. FAULT

87T

87G G

a) H.V. SIDE FAULT b) FAULT IN THE GENERATOR ZONE

0,4kV

c) FAULT IN THE 0.4kV SYSTEM

87T

87G G

FAULT IN THEGENERATOR ZONE

Fig. 3.2-6 Various fault points.


135

Fig. 3.2-7 Factor "k" as a function of time.


136

HEST 905 043 FL

A

I = 0

I

V V

1

20

220 VR 2 2

U = 20 x U2N + 20 x R2I2N

0

HEST 905 048 FL

2N

I2

E 2 2 2N(at I = 20 I )

E 2 (at I = I )2 2N

I �

0,01 x I20 2N

I �

0,05 x 20 I20 2N 2N= I

U2

Fig. 3.2-8 Measurement of the no-load characteristic of a current transformer.I1 = 0Influence of iron losses neglected.


137

Protectedmachine

Generator

Motor

Reactor

Transformer

CTcurrent error

Transformer no-load current

Tap changer range

10%

10% 10% 10%

10%-- --

Totalvalue

30% 30%

15% or20%

Typicalsetting g (%)

Table 3.2-I Component of the differential current at rated current.Typical values of the basic setting value "g".


138

HEST 905 019 FL

GT

G

I>

Z<

Protection zone

Fig. 3.3-1 Underimpedance protection


139

Z-setting

+0.07

-0.07

0

HEST 905 027 FL

x

z

r

�

Fig. 3.3-2 Operating characteristic of the underimpedance functionSetting: Z-Setting = 0.07Z = actual impedance


140

Generator

R S T

Voltagetransformer

Voltagerelay

Fig. 3.4-1 Voltage detection of generator interturn faults

R S T

Currentrelay

Currenttransformer

Neutral points

1

2

Fig. 3.4-2 Current detection of generator interturn faults


141

Neutral points

R S T

CurrentRelay

ct 1

Earthing resistor

ct 2

1

2

Fig. 3.4-3 Current detection of generator interturn faults


142

R S T

R1 S1 T1 R2 S2 T2

R S T

R1 0 R2 S1 0 S2 T1 0 T2

Differential relay

ct 1 ct 2 ct 3ct 4 ct 5 ct 6

Fig. 3.4-4 Differential current detection of generator interturn faults


143

Fig. 3.4-5 Interturn fault of generator stator windingmeasured voltagesa) Circuit breaker openb) Circuit breaker closed, generator connected to a big power system


144

I >

1

2

X

X

N N N1

N2 I2

I2

I1

N = N1 + N2

I1 = 0

N N N N N = N3 N1

N2

I1

I1

Fig. 3.4-6 Interturn fault of generator windingCirculating branch currents at circuit breaker open positiona) No winding branches connected in parellelb) Winding with 2 branches connected in parallelN1, N2, N3, N - number of winding turns


145

X

E

E1 X1

N1

I1

I2

X31

X13

N3 = N

X

X32

X23

E2

X12

X21 X2N2

I1

Fig. 3.4-7 Interturn fault of generator stator windingCircuit breaker open; 2 winding branches connected in parellelSelf and mutual reactancesN1, N2, N3, N - number of winding turns


146

Fig. 3.4-8 Interturn fault of generator stator windingCurrents i1, i2, and voltage u1 of the faulted branch as a function the number of shorted stator phase turns N2.Circuit breaker open; 2 winding branches connected in parellelN3 = N1 + N2 = N = 40


147

2

Feeder

1xsi1

i1

i1

xs

xs

rw1N2 - N2K

N2K i2K

T

Fig. 3.4-9 Interturn fault of transformer secondary windingCurrents i1 and i2sc as functions of the faulted winding length[x = (N2sc/N2)100%].Cicuit resistance not involved.Power system reactance xs = 0.005. Transformer reactance xT = 0.10.Variable leakage reactance: i1 - curve Nr. 1

i2sc - curve Nr. 2Constant leakage reactance i1 - curve Nr. 3

i2sc - curve Nr. 4


148

Fig. 3.4-10 Interturn fault of transformer secondary windingCurrents i1 and i2sc as functions of the faulted winding length[x = (N2sc/N2)100%].Cicuit resistance involved.Power system reactance xs = 0.005.Transformer reactance xT = 0.1, circuit resistance r = 0.01.Variable leakage reactance: i1 - curve Nr. 1


i2sc - curve Nr. 4


149

21xs i1

i1

i1

xs

xs

A

N1-N1KN1K

T

B

C

i1K

Fig. 3.4-11 Interturn fault of transformer primary windingCurrents i1 and i2sc as functions of the faulted winding length[x = (N1sc/N1)100%].Cicuit resistance not involved.Power system reactance xs = 0.005.Transformer primary winding leakage reactance x1� = 0.05.Variable leakage reactance: i1 - curve Nr. 1


i1sc - curve Nr. 4


150

Fig. 3.4-12 Interturn fault of transformer primary windingCurrents i1 and i1sc as functions of the faulted winding length[x = (N1sc/N1)100%].Cicuit resistance involved.Power system reactance xs = 0.005. Circuit resistance r = 0.01.Transformer primary winding leakage reactance x1� = 0.05.Variable leakage reactance: i1 - curve Nr. 1


i1sc - curve Nr. 4


151

G

UC UB

UAUCG UBG

UAG = 0

IE

IL

IC

IR

UE = UA

C LIL

IEIC

UA

UC UB

UE��Uph; IEmax

RE UE

IR

UE;IE

G(ground)

Fig. 3.6-1 Stator ground fault.Generator neutral point high-resistance grounded.


152

G

UC UB

UAUCG UBG

G

UA3 = 0 U3

I3UC3 UB3

IE

IL

IC

IR

C LIL IE

G

I3

U3

Re

IC

UA

UC UB

(ground)

UC3

Fig. 3.6-2 Stator ground fault.Generator terminal side high-resistance grounded.


153

IE>

C BRE

IEmax

A

Ground

Fig. 3.6-3 Stator ground fault protection applying a measurement of thegrounding resistance current.


154

Ic

Generator Generator transformer

Ic

3 C12Ic

3C12 Ic

IcCRE

U2ph

RE

vt

UE>

Ic

U2ph

1 2

Fig. 3.6-4 Influence of the H.V. ground fault on stator ground fault protectionfor generators operating with generator transformers.


155

REr

RES

RPr

RPS

REX 010

Injection Test

[V]

110

-110

[ms]Coded 12.5Hz injection Signal

UDC110V REG 216

AD

CPU

(95%) StEF Stator-EFP Rotor-EFP

UStator 95%

UStator 100%

URotor Ui

WU30Z

P8

REX 011

G

3 ~

N11

N12

US

Voltage

US Ur

UN=100V UN=25VUN=15VUN=100V

Injection Unit

InjectionTransformer Block

UiUirUis

Ugen/��100V

UiP

2*2 �F

+ - Ur Uir

110V 50V

110V

25.8V

Fig. 3.6-5 100% stator and rotor ground fault protection.


156

T 1

T 2

T 3

G 4

G 3

G 2

G 1

zone ofgenerator feeders

zone of the bus and of transformer feeders

Fig. 3.6-6 Generators directly connected to a busbar.


157

Generator Zero sequence ct

3 C12

UE>

UE>IE>

Rp

Re

Transformer

voltage transformer

Voltage transformer

Interposing VT

Fig. 3.6-7 Stator ground fault protection for generators directly connected to busbars


158

HEST 905 016 FL

Rotor

Rotor shaft

R

C1

50 V100 V

2CC

R

R R

50 V 100 VC1 2C

C

U

Fig. 3.7-1 Rotor ground fault protection


159

50% of Uph

UE >

UE

R

Fig. 3.8-1 Ground fault protection for ungrounded systems.


160

input

A

B

C

output

IE >

i 1

i 1

i 2

i 2Z E

x

0 10020 40 60 80

1,0

0,8

0,6

0,4

0,2

i 1 i 2

x [%]

80%

42%

i 1

i s

i 2

13

� I >

l

IE >

E >

Grounding impedance

ZE >> XT

Fig. 3.8-2 Restricted ground fault protection at impedance grounded transformerneutral point.a) grounding impedance determined for the maximum earth fault

current which is equal to the rated currentb) relay setting selected for 80% for the protected winding; this

means a setting value of is = i2 = 20%


161

RE

CT

RS

I >E

1 T 2

CT's

Fig. 3.8-3 Restricted ground fault protection of a delta - wye connected transformer.


162

IE>

Fig. 3.8-4 Transformer tank leakage protection.


163

trip signals

remote trip

I

I >

62timer50 BF

remote trip

trip signals

I 50 BF

releaseremote trip

timer62

I >

trip signalstimer62

C.B.

current

current

+

Fig. 3.9-1 Breaker failure protection schemes.


164

~ ~

Correct operationof B.F. protection

Maloperation of B.F. protection(C.B. is right and already open)

Fig. 3.9-2 Current detection at a breaker failure


165

t

t

fault clearing

delay by 62

50 BF

remotetrip signal

localC.B. closed open

time

faultcurrent

localtrip signal

setting level

C.B. operatingtime

0

reset time

margin

fault clearing

50 BF

remotetrip signal

localC.B. closed

time

faultcurrent

localtrip signal

setting level

0

delay by 62 remote C.B.operating time

remoteC.B. closed open

Fig. 3.9-3 Time chart of the device 50/62 BF.


166

fault clearing

localC.B. closed

time

faultcurrent

t

remotetrip signal

localtrip signal

0

remote C.B.operating time

remoteC.B. closed open

fault clearing

localC.B. closed open

time

faultcurrent

t

delay by 62

localtrip signal C.B. operating

time

0

reset time

margin

Fig. 3.9-4 Time chart of the device 62 BF


167�

0 t

i

i = iN

1.5

1.0�Nw = 100 ° C

�Noil = 90 ° C

��w = 10 min

��Nw = 60 K

��oil = 50 K

�oil = 120 min

i

20

40

60

80

100

120

140

160

0 20 40 60 80 100 120 140

�Nw

�Noil

�w��oil

�oil

��w��oil

t (min)

�oil

�

w

�w

�oil

(t =��(t =��

� (°C)

Fig. 4.1-1 Temperature rise at a sudden increased current


168

0 100 200 300 400 500

t (min)

10 100

110

120

110

120

130

140

winding temperature

simulated temperature rise

overload i = 1.2temperature rise atrated currentrelay setting of the thermal time constant

� ��C� ��

� = 90 min

��Nw � ��Noil= 10 K��Noil= 50 K �oil = 120 min

�w = 10 min

� = 126.4� C

Fig. 4.1-2 Comparison of an actual winding temperature rise and of a temperature rise simulated by a thermal relay


169�

Fig. 4.1-3 Temperature rise at a sudden current change.


170

Fig. 4.1-4 Short-time overload capability.


171�

Fig. 4.1-5 ABB stator overload device.Operating curve acc. to ANSI C50.13. specified by the following values:

IIN

2.26 1.54 1.3 1.16

t [s] 10 30 60 120


172

Fig. 4.2-1 Temperature rise at a sudden negative-sequence current change.


173�

0 t

Tripping stage

Alarm stage

HEST 905 015 FL

I2IN

I2IN

Fig. 4.2-2 Operating characteristics of the definite time NPS protection device.


174

Fig. 4.2-3 ABB device "NPS-Inverse" operating curve.


175�

A B C

IA IB IC

IA

IBIC

a2 IC

a IB

IA

3 I1 a2 IB

a IC

IA IA

IC

IB

3 I1 = 3 IA 3 I2 = 0 3 I0 = 0

I1 = IA I2 = 0 I0 = 0

Fig. 4.2-4 Symmetrical components at a symmetrical load.


176

IC

IA

IA IC

A CB

IA IC

IAIA

3 I1

3 I2

a2 IC

a IC

3 I0 = 0

I0 = 0

3 I2 = � 3 IA

I2 = � IA3333

3 I1 = � 3 IA

I1 = � IA33

Fig. 4.2-5 Symmetrical components at fully asymmetrical load.


177�

IB

IC

IA

IB IC

IA aIB

a2 IC

IA

a2 IBI2

IA

a IC

A CB

IA IB IC

3 I1 = 0

I1 = 0

3 I0 = 03 I2 = 3 IA

I2 = IA I0 = 0

Fig. 4.2-6 Symmetrical components at symmetrical load supplied by negative sequence load.


178

generatorline-to-line fault

unbalanced load orM.V. circuitbreaker fail. at IN

asymmetricalH.V. short circuits

I2IN

100 (%) 3331 ... 57.7

XX

200 167

X

H.V. circuit breaker failure at IN

100 (%) 57.7 33.3I2IN

XX

Table 4.2-I Values of the negative sequence current at various conditions.


179�

Generator type

Indirectly cooled rotor

air-cooled

hydrogen-cooled

Directly cooled rotor

... 350 MVA

350 ... 900 MVA

900 ... 1250 MVA

1250 ... 1600 MVA

Indirectly cooled

Directly cooled

Salient pole

generator

Turbo-

generator

10

10

15

10

8

8 .. 6

6 .. 5

5

8

8 .. 5

5

5

8

5

20

15

i2 �� [%] i22 t [s]

Table 4.2-II Generator capability at unbalanced stator currents according toIEC 34-1 / VDE 0530.


180

Generator type

Indirectly cooled rotor

Directly cooled rotor

... 960 MVA

961 ... 1200 MVA

1201 ... 1600 MVA

... 800 MVA

800 ... 1600 MVA

Salient pole

generator

10 30

8

6

5

10

10 ...5

i2�� [%] i22 t [s]

Table 4.2-III Generator capability at unbalanced stator currentsaccording to ANSI C50.13


181�

U set 2

U set 1

0 t

UN

Stage 2Stage 1

Delay 2 Delay 1

Fig. 4.3-1 Operating characteristic of the definite time overvoltage functionUN = rated voltage of the protection


182

G

EXd

Us

U

I

If

Ss = �

Us = constant

IfI = 0

E U Us

If

UE

I�

jXdI

If

jXdI U

E

�I

Fig. 4.6-1 Synchronous machine operating in parallel with a large power system.Reactive and capacitive power loading.


183�

Fig. 4.6-2 Current curves of a synchronous machine with xd = 2.0.


184

I

jXdIE

U

��

jXdI

E

jXqI

U

��

I

b) Salient-pole

generator

a) Turbo-

generator

Fig. 4.6-3 Generator phasors under normal operating conditions.


185�

XdI

�E

I

U

XdI

�E

I

U

Fig. 4.6-4 Synchronous machine with cylindrical rotorActive power loading.

PP

0 90 180 270 360 ��

Fig. 4.6-5 Active power of a synchronous generator as a function of the load angle �


186

jXdI

E

U

��

�

IIf

Xd Icos �

�E

UI

jXdI

P = U I cos �

Xd Icos �sin =�

E

P = UEXd

sin �

E = 0

U

I

jXdI

jXdI jXdI

EE�

UII

Fig. 4.6-6 Voltage diagram of a turbogenerator for different load conditions.


187�

G

EXd

Us

U

I

E EI

U UI

XdXdI

�

0R

Xd

EI

UI�

X

Fig. 4.6-7 Turbogenerator directly connected to a power system.Steady-state stability curve.


188

UXd

�0

Icmax Iactive

Iactive max.

Ireactive

IEXd

U = const

�

�

0

Qcmax P

P max.

Q

S=UIEUXd

U = const

�

UXd

Qexc=2

Fig. 4.6-8 Synchronous machine with cylindrical rotorPower diagram at constant stator voltage.


189�

0 R

Xd

EI

UsI

�

X

XT

GEXd XTU

I

Us

GT

jXTI

EU Us

�

jXdIjXTI

E

U Us

�

jXdI

Fig. 4.6-9 Generator with the generator transformer operating to a power system.Steady-state stability curve.


190

GEXd XTV

I

Us

GT

0 R

X

XT

X'd2

Loss-of-excitationdevice curve

Xd

UIZ =

Steady-statestability curve

Fig. 4.6-10 Coordination of steady-state stability curve with characteristic of the loss-of-excitation device.


191�

t

t

t

resetdelay

trip delay

extendedsignal

trippingsignal

protectiondevicestart signal

t

t

tresetdelay

trip delay

extendedsignal

trippingsignal

protectiondevicestart signal

trip

no trip

Fig. 4.6-11 Loss-of-excitation device.Trip at rotor swings.


192

�0

B' P

Q

S=UI

EUXd

U = const

�

UXd

2

B'

A'

jXdI U

I �

Point A, A':E = 0

I = UXd

��cap

I

E

��

U

jXdI

jXdIE

I

Point B:U = 0

I = EXd

��

R

U = Us = const

Xd

0B

X

Z

�

90°

G

E

Xd

Z= VI

jXdI

E

UI

��

A

Fig. 4.6-12a Turbogenerator directly connected to a large power system with a constant voltage.Steady-state stability limit (� = 90°)


193�

B'

U2

X’d2

U2

Xd

0 Q

P

A'

Us=U=const(XT=0)

EUXd

UI=S�

�

I = 2UX'd

-E = U

��90�cap��180°��

Point B':

I

jX'dI

�

U

E

I = UXd

E = 0

��90�cap��90��

I

Point A':

jXdI

�

U

R

Xd

0B

X

�

Z= UI

jXdI E

USI

�

A

jXTIU

�

UsG

E

Xd XT

U

T

X'd

X'd2

Z

Fig. 4.6-12b TurbogeneratorTypical device characteristic (90° < � < 180°).R-X and power diagram.


194

0A'

P

QB'

U=const.

EUXd

UXd

2 UXT

2

�G

S= UI

Point A':

jXdI

Us

I �

U

jXTI

I = UXd

E = 0��cap��G

Us =Xd+XT

XdU

Point B':

jXdIE

I�

U jXTI

Us = 0��90°ind� ��0°��G

E =Xd+XT

XTU

I= UXT

R

Us

Xd

0B

X

�

Z= UI

jXdI E

USI

��

A

G

E

Xd XT

U

T

XT

G��

jXTIU

�G

Fig. 4.6-13 Generator transformer unit.Steady-state stability limit (� = 90°)


195�

F3

B3

D3

E3

C2

C3

C1

No.1

No.2

No.3

A1=A2=A3

R0

B2D2

�

D1

B1

E2

B'1

B'2

E'2

D'3C'1C'3

F'3

No.1

No.2

No.3A'1=A'2=A'3

Q0

B'3

E'3

�

P

D'1

U=const

U2

XT

U2

Xd

D'2

Fig. 4.6-14 Application of the REG216/316 device "Minimum-reactance" at different load angle limits


196

�

Id

�

E

j Xq Iq

j Xd Id

U

Iq

I

�

Id

�

j Iq j Id

Iq

I

�

EXd

j ILXqXd

EXd

j IUXd

Xd - Xq UXd Xq

Fig. 4.6-15 Voltage and current diagram of a salient-pole generator.


197�

U2

Xd

EUXd

�

U2

Xq

0

P = P1+P2

Q

Xd - XqXd Xq

sin2�P2=U2

2

sin�P1=EUXdS = UI

P

�

�Id

�

j U IqjU Id

Iq

I

EUXd

EUXd

j U I

�

Xd - XqXd Xq

U2

U2

Xd

Fig. 4.6-16 Power diagram of a salient-pole generator.


198

Fig. 4.6-17 Steady-state stability limit for a salient-pole generator.———— salient-pole generator------------ turbo generator


199�

Fig. 4.6-18 Steady-state stability limit for salient-pole generator and generator-transformer units.———— salient-pole generator------------ turbogenerator (for comparison)


200

Fig. 4.6-19 Impedance locus after loss of excitation.


201�

Fig. 4.6-20 Power components after loss of excitation.


202

Fig. 4.6-21 Stator voltage, current and speed after loss of excitation.


203�

GEG

GT

network system

Xd XT

GEGU

G ES

XS

US

�

XT + XS UI

Xd EGI

ESI

�XTI U

XdI EG

ESXSI

�

UES

XdI

EG(XT + XS)I

Fig. 4.7-1 Generator with generator transformer operating to the network system under steady-state conditions.


204

EG

ES

��E

��E

EG

ES

EG

ES

��E

Fig. 4.7-2 Phasors EG and ES under 'out-of-step' conditions.


205�

EG > ES

EG < ES

0X'd

XS

XT

�R

EG = ES� �

X

Fig. 4.7-3 Impedance locus under out-of-step conditions.


206

Fig. 4.7-4 Locus of the impedance measured at the generator terminals during pole slipping in relation to the power system A.

X'd : transient reactance of the generatorXT : short-circuit reactance of the step-up transformerZS : transient impedance of the power system A


207�

X

0R

ES=0

EG=3.0

EG=1.6

EG=1.4

EG=1.3

EG=1.2

EG=1.8

2.0

3.0

1.0

xS

xT

x'd

Fig. 4.7-5 Impedance locus on the R-X diagram at ES = 1.0 and different voltages EG > ES.x'd = xT = xS = 0.2 (non-typical values).


208

X

R

ES=0.2

EG=0.4

EG=1.6

EG=0.7

EG=0.8

EG=0.85

EG=1.8

2.0

3.0

1.0

xS

xT

x'd

EG=ES=1.0

X'D = XT = XS = 0.2

Fig. 4.7-6 Impedance locus at ES = 1.0 and different voltages. EG < ES.- � - � - � - operating curve of 'loss-of-excitation' relay.


209�

Qcap 0 Qind

generator andturbine no-loadlosses

reversepower devicesetting

PM

PG

Fig. 4.8-1 Reverse-power device characteristic.


210

HEST 935 018 FL

U

IP>

Block

TRIP

Integrator

t >1

t >2

t >3

Fig. 4.8-2 Tripping circuit of a reverse power protection scheme forsteam turbines


211�

� = 300 ms�2 = 0.63662 (i.e. i0 = 0.5%)� = 314.16 1/1

Fig. 5.1-1 The factor k calculated according to 3 different formulas.


212

Fig. 5.1-2 The factor k as a function of the time t.


213�

Fig. 5.1-3 The factor km as a function of the time constant T2.

IMPORTANT NOTICE!

Experience has shown that reliable operation of our products isassured, providing the information and recommendations containedin these Instructions for Installation and Operation are adhered to.

It is scarcely possible for the instructions to cover every eventualitythat can occur when using technical devices and systems. Wewould therefore request the user to notify us directly or our agent ofany unusual observations or of instances, in which theseinstructions provide no or insufficient information.

In addition to these instructions, any applicable local regulationsand safety procedures must always be strictly observed both whenconnecting up and commissioning this equipment.

Any work such as insertion or removal of soldered jumpers orsetting resistors, which may be necessary, may only be performedby appropriately qualified personnel.

We expressly accept no responsibility for any direct damage, whichmay result from incorrect operation of this equipment, even if noreference is made to the particular situation in the Instructions forInstallation and Operation.

ABB Network PartnerABB Network Partner AGHaselstrasse 16/122CH-5401 Baden/SwitzerlandPhone +41 56 205 77 44Fax +41 56 205 55 77

ABB Network Partner ABS-72171 Västerås, SwedenPhone +46 21 32 13 00Fax +46 21 14 69 18

ABB Transmit OyRelays and Network ControlP.O.Box 699SF-65101 Vaasa, FinlandPhone +358-61-162 111Fax +358-61-161 094

ABB Power T&D Co.RelaysCoral Springs, Fla. 33065, USAPhone +1 305 752-6700Fax +1 305 752-6700, ext. 2283

Printed in Switzerland (9810-0000-0)

1MRB520046-Len

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